University  of  California  •  Berkeley 

The  Theodore  P.  Hill  Collection 

0/ 

Early  American  Mathematics  Books 


THE 


PEOGEESSIVE 


RACTICAL  ARITHMETIC, 

CONTAINING 

THE  TUEOUY  OF  NUMBERS,  IX  CONNECTION  WITH  CONCISE  ANALYTIC 

AND   SYNTHETIC  METHODS  OF   SOLUTION,  AND  DESIGNED 

AS  A  COMPLETE  TEXT  BOOK  ON  THIS  SCIENCE, 


FOR 


COMMOxN  SCHOOLS  AND  ACADEMIES. 


BY 

DANIEL  W.  FISH,  A.M., 

AUTUOE   or  THK  TABLE-BOOK,   PBIMAKT    AND   ISTELLEOTCTAL   AEITHMETICB,   AND 

RUDIMENTS. 


NEW  YORK : 
IVISON,  PHINNEY,  BLAKEMAN  &  CO., 

CHICAGO  :  S.  C.  GRIGGS  &  CO. 

1  864. 


RO  B  I  ISr  S  ON'S 

The  most  Complete,  most  Peactical,  and  most  Scientific  Series  of 
Mathematical  Text-Books  ever  issued  in  this  country. 


I.    Robinson's  Progressive  Table  Book, 

II.  Kobinson's  Progressive  Primary  Arithmetic,  -       -       -       - 

III.  Kobinson's  Progressive  Intellectual  Arithmetic,    -       -       - 

IV.  Robinson's  Rudiments  of  "Written  Arithmetic,      ... 
V.  Robinson's  Progressive  Practical  Arithmetic,        ... 

VI.    Robinson's  Key  to  Practical  Arithmetic, 

VII.  Robinson's  Progressive  Higher  Arithmetic,    -       -       -       - 

VIII.    Robinson's  Key  to  Higher  Arithmetic, 

IX.    Robinson's  New  Elementary  Algebra, 

X.    Robinson's  Key  to  Elementary  Algebra, 

XI.    Robinson's  University  Algebra, 

XII.    Robinson's  Key  to  University  Algebra, 

XIII.  Robinson's  New  University  Algebra, 

XIV.  Robinson's  Key  to  New  University  Algebra, ...       - 
XV.  Robinson's  New  Geometry  and  Trigonometry,    - 

XVI.    Robinson's  Surveying  and  Navigation, 

XVII.  Robinson's  Analyt.  Geometry  and  Conic  Sections,      - 

XVIII.  Robinson's  Differen.  and  Int.  Calculus,  (in  preparation,). 

XIX.    Robinson's  Elementary  Astronomy, 

XX.    Robinson's  University  Astronomy, 

XXI.    Robinson's  Mathematical  Operations, 

XXII.  Robinson's   Key  to  Geometry  and  Trigonometry,  Conic 

Sections  and  Analytical  Geometry, 


Entered,  accordiii;,'  to  Act  of  Congress,  in  the  year  IS58,  by 

IIOKATIO    N.    IIOBINSON,    LL.D., 

and  again  in  tlie  year  1SG3,  by  - 

DANIEL    W.    FISH,    A.M., 

In  the  Clerk's  OflQco  of  tlie  District  Court  of  tlio  United  States,  for  the  Northern 

District  of  Now  York. 


PREFACE. 


Progress  and  improvement  characterize  almost  every  art  and 
science ;  and  within  the  last  few  years  the  science  of  Aritlimetic  has 
received  many  important  additions  and  improvements,  which  have 
appeared  from  time  to  time  successively  in  the  different  treatises  pub- 
lished upon  this  subject. 

In  the  preparation  of  this  work  it  has  been  the  author's  aim  to  com- 
bine, and  to  present  in  one  harmonious  whole,  all  these  modern  im- 
provements, as  well  as  to  introduce  some  new  methods  and  practical 
operations  not  found  m  other  works  of  the  same  grade ;  in  short,  to 
present  the  subject  of  Arithmetic  to  the  pupil  more  as  a  science  than 
an  art;  to  teach  him  methods  of  thought,  and  how  to  reaso7i,  rather 
than  xchat  to  do;  to  give  unity,  system,  and  practical  utility  to  the 
science  and  art  of  computation. 

The  author  believes  that  both  teacher  and  pupil  should  have  the 
privilege,  as  well  as  the  benefit,  of  performing  at  least  a  part  of  the 
thinkincj  and  the  labor  necessary  to  the  study  of  Arithmetic ;  hence 
the  present  work  has  not  been  encumbered  with  the  multiplicity  of 
•'notes,"  "suggestions,"  and  superfluous  operations  so  common  to 
most  Practical  Aritlmictics  of  the  present  day,  and  which  prevent  the 
cultivation  of  that  self-reliance,  that  clearness  of  thought,  and  that 
vigor  of  intellect,  which  always  characterize  the  truly  educated  mind. 

The  author  claims  for  this  treatise  improvement  upon,  if  not  superi- 
ority over,  others  of  the  kind  in  the  following  particulars,  viz. :  In 
the  mechanical  and  typographical  strjle  of  the  icork ;  the  open  and 
attractive  page ;  the  progressive  and  scientific  arrangement  of  the 
subjects ;'  clearness  and  conciseness  of  definitio7is ;  fullness  and  accu- 
racy in  the  i}eio  and  improved  methods  of  operations  and  analyses; 
brevity   and  perspicuity   of  rules ;    and   in   the    very   large   number   of 

Ciii) 


iV  PREFACE. 

examples  prepared  and  arranged  tcith  special  reference  to  their  prac- 
tical utility,  and  their  adaptation  to  the  real  business  of  active  life. 
The  answers  to  a  part  of  the  examples  have  been  omitted,  that  the 
learner  may  acqtiire  the  discipline  resulting  from  verifj-ing  the  opera- 
tions. 

Particiilar  attention  is  in\'ited  to  improvements  in  the  subjects  of 
Common  Divisors,  Multiples,  Fractions,  Percentage,  Interest,  Propor- 
tion, Analj'sis,  Alligation,  and  the  Roots,  as  it  is  believed  these 
articles  contain  some  practical  features  not  common  to  other  authors 
upon  these  subjects. 

It  is  not  claimed  that  this  is  a  perfect  "work,  for  perfection  is  impossi- 
ble ;  but  no  effort  has  been  spared  to  present  a  clear,  scientific,  com- 
prehensive, and  complete  system,  sufficiently  full  for  the  business  man 
and  the  scholar ;  not  encumbered  with  unnecessary  theories,  and  yet 
combining  and  systematizing  real  improvements  of  a  practical  and 
useful  natujre.  How  nearly  this  end  has  been  attained  the  intelligent 
and  experienced  teacher  and  educator  must  determine. 


CONTENTS. 


SIMPLE     NUMBERS. 


PAGr 

Definitions, T 

Roman  Notation, 8 

Table  of  Pwoman  Notation, 9 

Arabic  Notation, 10 

N umeiation  Tabic, 15 

Laws  and  Rules  for  Notation  and 

Numeration, 17 

Addition, 20 

Subtraction, 28 

Multiplication, S5 

Contractions, 42 


PAGE 

Di  vision, 4T 

Contractions, 68 

Applications  of  preceding  Rules, ....  60 

General  Principles  of  Division, Gi 

Exact  Divisors, C6 

Piiuie  Numbers, CT 

Factoring  Numbers, 67 

Cancellation, 69 

Greatest  Common  Divisor, 73 

Multiples 79 

Classification  of  Numbers, 8-i 


COMMON    FRACTIONS. 


Definitions,  &c., 86 

Gener.al  Principles  of  Fractions, 89 

Reduction  of  Fractions, 83 

Addition  of  Fractions, 96 


Subtraction  of  Fractions, 93 

Multiplication  of  Fractions, 101 

Division  of  Fractions, 106 

Promiscuous  Examples, 112 


DECIMALS, 


Decimal  Notation  and  Numeration,.  116 

Reduction  of  Decimals, 121 

Addition  of  Decimals, 134 


Subtraction  of  Decimals, 126 

Multiplication  of  Decimals, 127 

Division  of  Decimals, 123 


DECIMAL     CtJRRENCT. 


Notation  and  Numeration  o.  Decimal 

Currency, 131 

Reduction  of  Decimal  Currency, 132 

Addition  of  Decimal  Currency, 134 

Subtraction  of  Decimal  Currency,. . .  185 
Multiplication  of  Decimal    Curren- 
cy   136 

Division  of  Decimal  Currency, 137 

Additional  Applications, 139 


"When  the  Price  is  an  Aliquot  Part 

of  a  Dollar 139 

To  find  the  Cost  of  a  Quantity, 140 

To  find  the  Price  of  One, 141 

To  find  the  Quantity, 141 

Articles  sold  by  the  100  or  1000, 143 

Articles  sold  by  the  Ton, 143 

Bills, 144 

Promiscuous  Examples, 148 


Tl 


CONTENTS. 


COMPOUND     NUMBERS. 


PAGE 

Eeduction, 150 

DeflnitioDS.  «fec., 150 

English  Money, 151 

Troy  Weight, 153 

Apothecaries'  Weight, 154 

Avoirdupois  Weight, 155 

Long  Measure, 15S 

Surveyors'  Long  Measure, KiO 

Square  Measure, ICl 

Surveyors'  Square  Measure, 164 

Cubic  Measure, 1C5 

Liquid  Measure, 1G7 

Dry  Measure, 1 03 

Time,..- 170 

Circular  Measure, 172 


PAOB 

Counting;  Paper;  Books,  Ac, 173 

lieduction  of  Denominate  Fractions,  175 
Addition  of  Compound  Numbers,. ..  183 
Addition  of  Denominate  Fractions, .  185 

Subtraction, 186 

To  find  the  Difference  in  Dates, 1S8 

Table, 1S9 

Subtraction    of   Denominate   Frac- 
tions,    190 

Multiplication  of  Compound  Num- 
bers,   191 

Division,...  193 

Longitude  and  Time, 195 

Duodecimals, 193 

Promiscuous  Examples, 202 


PERCENTAGE. 


Definitions,  &c., 205 

Com  mission  and  Brokerage, 210 

Stocks, 214 

Profit  and  Loss, 217 

Insurance, 223 

Taxes, 224 

Custom  House  Business, 227 

Simple  Interest, 230 


Partial  Payments  or  Indorsement.',..  -37 

Problems  in  Interest, 243 

Com])ound  Interest, 24G 

Discount, 249 

Banking, 252 

Exchange, , 256 

Equation  of  Payments, 201 


RATIO     AND     PROPORTION, 


Patio, 209 

Proportion, 272 

Simple  Proportion, 273 

Compound  Proportion, 279 

Partnership, 2S1 

An.-ilysis, 2SS 

Alligation  Medial, 297 

Alligation  Alternate, 293 


Involufion, Sfl.T 

Evolution, 304 

Square  Root, 305 

Cube  Root, 812 

Arithmetical  Progres.sion, 313 

Geometrical  Progression, 821 

Promiscuous  E.xamples, 824 

Mensuration ....,  ..  832 


PHACTICAL   ARITHMETIC. 


DEFINITIONS. 


1.  Quantity  is  any  thing  that  can  be  increased,  diminished, 

or  mca'^ured. 

S.    Mathematics  is  the  science  of  quantity. 

3.  A  Unit  is  one,  or  a  single  thing. 

4.  A  Number  is  a  unit,  or  a  collection  of  units.  , 

5.  An  Integer  is  a  whole  number. 

6.  The  Unit  of  a  Number  is  one  of  the  same  kind  or 
name  as  the  number.  Thus,  the  unit  of  23  is  1 ;  of  23  dollars, 
1  dollar  ;  of  23  feet,  1  foot. 

7.  Like  -Numbers  have  the  same  kind  of  unit.  Thus,  74, 
16,  and  250  ;  7  dollars  and  62  dollars  ;  19  pounds,  320  pounds, 
and  86  pounds;  4  feet  6  inches,  and  17  feet  9  inches. 

8.  An  Abstract  Number  is  a  number  used  without  refer- 
ence to  any  particular  tliiug  or  quantity.     Thus,  17;  365;  8540. 

9.  A  Concrete  Number  is  a  number  used  with  reference 
to  some  particular  thing  or  quantity.  Thus,  17  dollars;  365 
days ;  8540  men. 

XoTES.    1.    The  unit  of  an  abstract  number  is  1,  and  is  called  Trmfij. 

2.  Concrete  numbers  are,  by  some,  called  Denominate  Numbers- 
Denomination  means  the  name  o"f  the  unit  of  a  concrete  number. 

10.  Arithmetic  is  the  Science  of  numbers,  and  the  Art  of 
computation. 

11.  A  Sign  is  a  character  indicating  an  operation  to  be 
performed. 

12.  A  Rule  is  a  prescribed  method  of  performing  an  op- 
eration. 

Define  quantity.     Mathematics.     A  imit.     A  number.  An  integer. 

The  unit  of  a  number.     Like   numbers.     An   abstract  number.     A 

concrete   number.     The   unit  of  an   abstract  number.  Denominate 
numbers.     Arithmetic.     A  sign,  or  sjTnbol.     A  rule. 


8  SIMPLE   NUMBERS. 


NOTATION  AND   NUMERATION. 

IS.  Notation  is  a  method  of  z^n'^jHy  or  expressing  numbers 
by  charactei's ;  and, 

1 4: .  Numeration  is  a  method  of  reading  numbers  expressed 
by  characters. 

15.  Two  systems  of  notation  are  in  general  use  —  the 
Roman  and  the  Arabic. 

Note.  The  Roman  Notation  is  supposed  to  have  been  first  used 
by  the  Romans  ;  hence  its  name.  The  Arabic  Notation  was  intro- 
duced into  Europe  by  the  Arabs,  by  -whom  it  was  supposed  to  have 
been  invented.  Eut  investigations  have  shown  that  it  was  adopted  by 
them  about  600  years  ago,  and  that  it  has  been  in  use  among  the  Hin- 
doos more  than  2000  years.  From"  this  latter  fact  it  is  sometimes 
called  the  Indian  Notation. 

THE    ROMAN    NOTATIOX 

16.  Employs  seven  capital  letters  to  express  numbers, 
thus  : 


Letters,         I  V  X  L  C  D  M 

one'  five  one 

hundred,  buudred,  thoasand. 


AAoli^r^c  ,.  i  re  ona'  five  one 

values,  one,  five,  ten,  fifty, 


17,  The  Roman  notation  is  founded  upon  five  principles, 
as  follows : 

1st.  Repeating  a  letter  repeats  its  value.  Thus,  II  repre- 
sents two,  XX  twenty,  CCC  three  hundred, 

2d.  If  a  letter  of  any  value  be  plaeed  after  one  of  greater 
value,  its  value  is  to  be  united  to  that  of  the  greater.  Thus, 
XI  represents  eleven,  LX  sixty,  DC  six  hundred. 

3d.  If  a  letter  of  any  value  be  placed  before  one  of  greater 
value,  its  value  is  to  be  taken  from  that  of  the  greater.  Thus, 
IX  represents  nine,  XL  forty,  CD  four  hundred. 

Define  notation.  Numeration.  ^Miat  systems  of  notation  are  now 
in  general  use  ?  From  what  are  their  names  derived  ?  What  are  used 
to  express  numbers  in  the  Roman  notation  r  "What  is  the  value  of  each  ? 
What  is  the  first  principle  of  combination  ?     Second  ?     Third  ? 


NOTATION   AND   NUJIERATION. 


9 


4th.  If  a  letter  of  any  value  be  placed  hetiveen  two  letters, 
each  of  greater  value,  its  value  is  to  be  taken  from  tlie  united 
value  of  the  other  two.  Thus,  XIV  represents  fourteen, 
XXIX  twenty-nine,  XCIV  ninety-four. 

oth.  A  bar  or' dash  placed  over  a  letter  increases  its  value 
one  thousand  fold.  Thus,  V  signifies  five,  and  V  five  thou- 
sand ;  L  fifty,  and  L  fifty  thousand. 


TABLE    OF    KOMAN    NOTATION. 


I 

is 

One. 

II 

Two. 

III 

Three. 

IV 

Four. 

V 

Five. 

VI 

Six. 

VII 

Seven. 

VIII 

Eight. 

IX 

X^ine. 

X 

Ten. 

XI 

Eleven. 

XII 

Twelve. 

XIII 

Thirteen. 

XIV 

Fourteen. 

XV 

Fifteen. 

XVI 

Sixteen. 

XVII 

Seventeen 

^VIII 

Eigliteen. 

XIX 

X'ineteen. 

XX 

is   Twenty. 

XXI 

"  Twenty-one. 

XXX 

"    Thirty. 

XL 

"   Forty. 

L 

"    Fifty. 

LX 

"    Sixty. 

LXX 

"    Seventy. 

LXXX 

"   Ei<>:hty. 

xc 

"   Ninety. 

c 

"   One  hundred. 

cc 

"    Two  hundred. 

D 

"    Five  hundred. 

DC 

"    Six  hundred. 

M 

"   One  thousand. 

[dred 

MC 

"   One  thousand 

one  hun- 

MM 

"   Two  thousand. 

x 

"    Ten  thousand. 

C 

"    One  hundred  thousand 

M 

"    One  million. 

NoTE.     The  system  of  Roman  notation 


p\n-poses  of  niimcrical  calculation ;    it  is  principally  confined  to  the 


s  not  well  adapted  to  the 


numbering  of  chapters  and  sections  of  book 

Express  the  fjllowing  numbers  by  letters : 

1.  Eleven. 

2.  Fifteen. 


piiblic  docmnents,  &c. 


Ans. 
AnS' 


XL 


Fourth  ?     Fifth  ?     Repeat  the  table.     ^Miat  is  the  value  of  I-VTI  ? 
CLXXIII?  XCVin? CDXXXII?  XCIX?  DCXIX? 

V-MDCCXLIX?      MDXXVCDLXXXIX  ?      To    what     uses    is    the 
Roman  notation  now  principally  confined  ? 
1* 


10  SIMPLE    NUMBERS. 

3.  Twenty-five. 

4.  Thirty-nine. 

5.  Forty-eight. 

6.  Seventy-seven. 

7.  One  hundred  fifty-nine. 

8.  Five  hundred  ninety-four. 

9.  One  thousand  five  hundred  thirty-eight. 

10.  One  thousand  nine  hundred  ten. 

11.  Express  the  present  year. 

THE    ARABIC    NOTATION 

IS.    Employs  ten  chai-acters  or  figures  to  express  numbers. 
Thus, 
Figures,  012345G789 

Names  and  ')    "^^"S^*    °^^'     *^^'°j    three,   four,     five,      Bix,     Beven,    eight,    nine, 
values,         5  ^i,,°'„^ 


19.  The  first  character  is  called  naught,  because  it  has  no 
value  of  its  own.  The  other  nine  characters  are  called  signif- 
icant figures,  because  each  has  a  value  of  its  own. 

20.  The  significant  figures  are  also  called  Digits,  a  word  de- 
rived from  the  Latin  terra  digitus,  which  signifies ^;(^er. 

21.  The  naught  or  cipher  is  also  called  nothing,  and  zero. 

The  ten  Arabic  characters  are  the  Alphabet  of  Aritlimetic, 
and  by  combining  tliem  according  to  certain  principles,  all 
numbers  can  be  expressed.  We  will  now  examine  the  most 
important  of  these  princii)les.* 

92.  Each  of  tlie  nine  digits  has  a  value  of  its  own  ;  hence 
any  number  not  greater  than  9  can  be  expressed  by  one 
figure. 


*  Fractional  and  Jecimtil  not.-ition,  and  the  notation  of  compound  numbers,  will  bo 
discussed   .1  their  appropriate  places. 

"\Miat  are  used  to  cxpro'ss  nunibtTs  in  the  Arabic  notation  ?  '^^'ll,^t 
is  the  value  of  each  ?  What  general  name  is  given  to  the  signiticant 
figures  ?     AVhy  ?     Numbers  less  than  ten,  how  c.xiivessed  ? 


NOTATION   AND   NUMERATION.  H 

S3.  As  we  have  no  single  character  to  represent  ten,  we 
express  it  by  writing  the  unit,  1,  at  the  left  of  the  cipher,  0, 
thus,  10.     In  the  same  manner  we  represent 

2  tens,  3  tens,  4  tens,  5  tens,  G  tens,  7  tens,  8  tons,  9  tens, 

or  or             or          or  or  or  or                or 

twenty,  thirty,  forty,     fifty,  sixty,  seventy,  ciglity,  ninety, 

20;  30;        40;       50;  GO;  70;  80;          90. 

S4:.  When  a  number  is  expressed  by  two  figures,  the  right 
hand  figure  is  called  units,  and  the  left  hand  figure  tens. 

We  express  the  numbers  between  10  and  20  by  Avriting 
the  1  in  the  place  of  tens,  with  each  of  the  digits  respectively 
in  the  place  of  units.     Thus, 

eleven,    twelve,    thirteen,  fourteen,    fifteen,   sixteen,  seventeen,  eighteen,  nineteen. 

11,       12,       13,       14,        15,      16,         17,         18,        19. 

In  like  manner  we  express  the  numbers  between  20  and 
30,  between  30  and  40,  and  between  any  two  successive  tens. 
Thus,  21,  22,  23,  24,  25,  26,  27,  28,  29,  34,  47,  56,  72,  93. 
The  greatest  number  that  can  be  expressed  by  two  figures 
is  99. 

25.  We  express  one  hundred  by  writing  the  unit,  1,  at  the 
left  hand  of  two  ciphers,  or  the  number  10  at  the  left  hand  of 
one  cipher;  thus,  100.  .In  like  manner  we  write  two  hun- 
dred, three  hundred,  &c.,  to  nine  hundred.     Thus, 

one       t^\•o       three      four        five  six       seven     eight      nine 

hundred,  hundred,  hundred,  liun<Ired,  hundred,  hundred,  hundred,  hundred,  hundred, 

100,     200,      300,     400,     500,      GOO,      700,     800,     900. 

26.  When  a  number  is  expressed  by  three  figures,  the 
right  hand  figure  is  called  Slits,  the  second  figure  tens,  and 
the  left  hand  figure  hundreds. 

As  the  ciphers  have,  of  themselves,  no  value,  but  are  always 
used  to  denote  the  absence  of  value  in  the  places  they  occupy, 

Tens,  how  expressed  ?  The  right  hand  figure  called  what  -•  Left 
hand  figure,  wliat  ?  What  is  the  greatest  number  that  can  be  expressed 
by  two  figures  ?  One  hundred,  how  expressed  ?  A\Tien  numbers  are 
expressed  by  three  figures,  what  names  are  given  to  each  ? 


12  SIMPLE   NUMBERS. 

we  express  tens  and  units  with  hundreds,  by  writing,  in  place 
of  the  ciphers,  the  numbers  representing  the  tens  and  units. 
To  express  one  hundred  fifty  we  write  1  hundred,  5  tens,  and 
0  units ;  thus,  150.  To  express  seven  hundred  ninety-two, 
we  write  7  hundreds,  9  tens,  and  2  units  ;  thus, 

m 

TS 

'a  •  to 

3  P  £3 

7      9      2 

The  greatest  number  that  can  be  expressed  by  three  figures 
is  999. 

EXAMPLES   FOR    PRACTICE. 

1.  Write  one  hundred  twenty-five. 

2.  "Write  four  hundred  eighty-three. 

3.  "Write  seven  hundred  sixteen. 

4.  Express  by  figures  nine  hundred. 

5.  Express  by  figures  two  hundred  ninety. 

6.  "Write  eight  hundred  nine. 

7.  "Write  five  hundred  five. 

8.  "V\^rite  five  hundred  fifty-seven. 

97.  "We  express  one  thousand  By  writing  the  unit,  1,  at 
the  left  hand  of  three  ciphers,  the  number  10  at  the  left  hand 
of  two  ciphers,  or  the  number  100  at  the  left  hand  of  one 
cipher;  thus,  1000.  In  the  same  manner  we  write  two 
thousand,  three  thousand,  &c.,  to  nine  thousand ;  thus, 

one         two       three       four       five        six         seven     eight       nine 

thousand,  thniisand,  thousand,  thousand,  thousand,  thousand,  thousand,  thousand,  thousand, 

1000,  2000,  3000,   4000,  5000,  GOOO,  7000,   8000,  9000. 

38.  "When  a  number  is  expressed  by  four  figures,  the 
places,  commencing  at  the  right  hand,  are  units,  tc7is,  hundreds, 
thousands. 

Use  of  the  cipher,  what  ?  Greatest  number  that  can  be  expressed  by- 
three  figures  ?  One  thousand,  how  expressed  ?  How  many  figures 
used  ?     Names  of  each  ? 


NOTATION   AND  NUMERATION.  13 

To  express  hundreds,  tens,  and  units  with  thousands,  we 
write  in  each  place  the  hgure  indicating  tlie  number  we  wish 
to  express  in  that  place.  To  write  ibur  thousand  two  hun- 
dred sixty-nine,  we  write  4  m  the  place  of  thousands,  2  in  the 
place  of  hundreds,  6  in  the  place  of  tens,  and  9  in  the  place  of 
units ;  thus, 


:3 


t3 


4      2       6 


2       3       « 


The  greatest  number  that  can  be  expressed  hj  four  figures 
is  9990. 

EXAMPLES    FOR   PRACTICE. 

Express  the  following  numbers  by  figures  :  — 

1.  One  thousand  two  hundred. 

2.  Five  thousand  one  hundred  sixty. 

3.  Three  thousand  seven  hundred  forty-one. 

4.  Eight  thousand  fifty-six. 

5.  Two  thousand  ninety. 

6.  Seven  thousand  nine. 

7.  One  thousand  one. 

8.  Nine  thousand  four  hundred  twenty-seven. 

9.  Four  thousand  thirty-five. 

10.  One  thousand  nine  hundred  four. 
Read  the  following  numbers :  — 

11.  76;      128;      405;      910;      116;    3416;    1025. 

12.  2100;    5047;   7009;   4670;   3997;    1001. 

2@.  Next  to  thousands  come  tens  of  thousands,  and  next 
to  these  come  hundreds  of  thousands,  as  tens  and  hundreds 
come  in  their  order  after  units.  Ten  thousand  is  expressed 
by  removing  the  unit,  1,  one  place  to  the  left  of  the  place 


Greatest  number  expressed  by  four  figiires  ?     Tens  of  thousands,  how 
expressed  ?     Hundreds  of  thousands  ? 


14  SIMPLE   NUMBERS. 

of  thousands,  or  by  writing  it  at  the  left  hand  of  four  ci- 
phers; thus,  10000;  and  one  hundred  thousand  is  expressed 
by  removing  the  unit,  1,  still  one  place  further  to  the  left,  or 
by  writing  it  at  the  left  hand  of  five  ciphers ;  thus,  100000. 
We  can  express  thousands,  tens  of  thousands,  and  hundreds  of 
thousands  in  one  number,  in  the  same  manner  as  we  express 
units,  tens,  and  hundreds  in  one  number.  To  express  five 
hundred  twenty-one  thousand  eight  hundred  three,  we  write  5  in 
the  sixth  place,  counting  from  units,  2  in  the  fifth  place,  1  in 
the  fourth  place,  8  in  the  third  place,  0  in  the  second  place, 
(because  there  are  no  tens,)  and  3  in  the  place  of  units; 
thus, 


O 


O   C  PI 

Ci  rr.    ^  C3 


r— f      ^J  r-      ^  «J  -i^  • 

T3r-«  Cf^  r^  TJ  •  (71 

_g^  ^       ^         ^         o         S 

rC;    +-»  -*-t  4_»  rC5  ^J  P 

5         2        18      0      3 

The  greatest  number  that  can  be  expressed  by  five  figures 
is  99999  ;  and  by  six  figures,  999999. 

EXAMPLES    FOR    PRACTICE. 

"\Yi-ite  the  following  numbers  in  figures :  — 

1.  Twenty  thousand. 

2.  Forty-seven  thousand. 

3.  Eighteen  thousand  one  hundred. 

4.  Twelve  thousand  three  hundred  fifty. 

5.  Thirty-nine  thousand  five  hundred  twenty-two. 
G.  Fifteen  thousand  two  hundred  six. 

7.  Eleven  thousand  twenty-four. 

8.  Forty  thousand  ten, 

9.  Sixty  thousand  six  hundred. 
10.  Two  hundred  twenty  thousand. 

'     11.    One  hundred  fifty-six  tliousand. 

12.    Plight  hundred  forty  thousand  three  hundred. 


Greatest  number  expressed  by  five  figures  ?     Six  figures  ? 


NOTATION   AND   NUMEIIATION.  15 

13.  Five  hundred  one  thousand  nine  hundred  sixtj-four. 

14.  One  hundred  thousand  one  hundred. 

15.  Throe  hundred  thirteen  thousand  three  hundred  thir- 
teen. 

16.  Seven  liundred  eighteen  thousand  four. 

17.  One  hundn.'d  tliousand  ten. 
Read  the  following  numbers  :  — 

18.  5006;        12304;     96071;  5470;     203410. 

19.  36741;      400560;      13061;         49000;     100010. 

20.  200200;        75620;     90402;      218094;     lOOlOl. 

For  convenience  in  reading  large  numbers,  we  may  point 
them  off,  by  commas,  into  periods  of  three  figures  eaeh,  count- 
ing from  the  right  hand  or  unit  ligure.  This  pointing  enables 
us  to  read  the  hundreds,  tens,  and  units  in  each  period  with 
facility. 

SO.  Next  above  hundreds  of  thousands  we  have,  succes- 
sively, units,  tens,  and  hundreds  of  millions,  and  then  follow 
units,  tens,  and  hundreds  of  each  higher  name,  as  seen  in  the 
following 

NUMERATION   TABLE. 


• 

• 

1 

0 

a 

o 

, 

, 

■^3 

C 

•r-i 

•4— • 

.2 
3 

ja 
■^ 

u 
a 

tc 

.2 

o 

.2 

S 

. 
to 

a, 

to 

3 

n 
a 

s 

1 

v< 

«« 

«« 

«M 

Vl 

e« 

tw 

Ct-I 

tM 

O 

O 

o 

o 

o 

O 

O 

O 

o 

'^  i~i  j~*  u  '—  t^  u  ^ 

co'O  tn  "^  (r'3  m'D  rrT;  (eT3  tn  ^  m   -z  tn 

c  c  3  c  c  P  =  "=  5   c  ■::  5   c  •=  3  c  •=  3  c  'S  5   c  ^  5  c  "S 


9  8,765,432,109,876,556,789,012,345 

ninth  eiijhth  seventh   sixth      fifth      f. nirfh     tliird     second     first 
period,  period,  period,  iieriod.  period,  period,  period,  period,  period. 

IIow  may  fignres  be  pointed  off?  One  million,  how  expressed? 
^Next  period  above  millions,  what  ?  Give  the  name  of  each  successive 
period. 


16  SIMPLE   NUMBERS. 

XoTE.     This  is  called  the  French  method  of  pointing  off  the  peri- 
ods, and  is  the  one  in  general  use  in  this  country. 


Figures  occupying  diiferent  places  in  <i  number,  as 
units,  tens,  Lundreds,  &;c.,  are  said  to  express  different  orders 

of  units. 

Simple  units  are  called  units  of  the  ^rsf     order. 

Tens  "       "  "      "    "    second     " 

Hundreds  "       «  "      "    "    (/lird       « 

Thousands  "       "  "      «    "  fourlh  '  « 

Tens  of  thousands  "       "  "      "    "  ffth        « 

and  so  on.  Thus,  452  contains  4  units  of  the  third  order,  5 
units  of  the  second  order,  'and  2  units  of  the  first  order. 
1,030,000  contains  1  unit  of  the  seventh  order,  (millions.)  3 
units  of  the  fifth  order,  (tens  of  thousands.)  and  6  units  of  the 
third  order,  (hundreds.) 

EXAMPLES    FOR    PRACTICE. 

lYrite  and  read  the  following  numbers  :  — 

1.  One  unit  of  the  third  order,  four  of  the  second. 

2.  Three  units  of  the  fifth  order,  two  of  the  third,  one  of  the 
first. 

3.  Eight  units  of  the  fourth  order,  five  of  the  second. 

4.  Two  units  of  the  seventh  order,  nine  of  the  sixth,  four 
of  the  third,  one  of  the  second,  seven  of  the  first. 

5.  Three  units  of  the  sixth  order,  four  of  the  second. 

G.  Nine  units  of  the  eighth  order,  six  of  the  seventh,  three 
of  the  fifth,  seven  of  the  fourth,  nine  of  the  first. 

7.  Four  units  of  the  tenth  order,  six  of  the  eighth,  four  of 
the  seventh,  two  of  the  sixth,  one  of  the  third,  five  of  the  sec- 
ond. 

8.  Eiglit  units  of  the  twelfth  order,  four  of  the  eleventh,  six 
of  the  (cntli,  nine  of  the  scvontli,  three  of  the  sixth,  five  of  the 
fifth,  two  of  tiie  tliird,  eight  of  the  first. 

Units  of  dilFcrent  orders  are  ^vhat  i 


NOTATION   AND   NUMERATION.  17 

32,  From  the  foregoing  explanations  and  illustrations,  we 
derive  several  important  principles,  which  we  will  now  pre- 
sent. 

1st.  Figures  have  two  values.  Simple  and  Local. 

The  Simple  Value  of  a  figure  is  its  value  when  taken  alone; 
thus,  2,  5,  8. 

The  Local  Value  of  a  figure  is  its  value  when  used  with  an- 
other figure  or  figures  in  the  same  number ;  thus,  in  842  tl;e 
simple  values  of  the  several  figures  are  8,  4,  and  2;  but  the 
local  value  of  the  8  is  800  ;  of  the  4  is  4  tens,  or  40  ;  and  of 
the  2  is  2  units. 

Note.  When  a  figure  occupies  units'  place,  its  simple  and  local 
values  are  the  same. 

2d.  A  digit  or  figure,  if  used  in  the  second  place,  expresses 
tens  ;  in  the  third  jJace,  hundreds ;  in  the  fourth  place,  thou- 
sands ;  and  so  on. 

3d.  As  10  units  make  1  ten,  10  tens  1  hundred,  10  hun- 
dreds 1  thousand,  and  10  units  of  any  order,  or  in  any  place, 
make  one  unit  of  the  next  higher  order,  or  in  the  next  place 
at  the  left,  we  readily  see  that  the  Arabic  method  of  notation 
is  based  upon  the  following 

TWO    GENERAL    LAWS. 

T.  Th3  different  orders  of  units  increase  from  right  to  left, 
and  decrease  from  left  to  right  ^  in  a  tenfold  ratio. 

II.  Every  removal  of  a  figure  one  place  to  the  left,  increases 
its  local  value  tenfold  ;  and  every  removal  of  a  figure  one  2dace 
to  the  right,  diminishes  its  local  value  tenfold. 

Thus, 

6  is  6  units. 

60  is  10  times  6  units. 

600  is  10  times  6  tens. 

6000  is  10  times  6  hundreds. 

COOOO  is  10  times  6  thousands. 


First  principle  derived  ?    "What  is  the  simple  value  of  a  figure  ?    Local: 
Secoiid  principle  ?     Third  ?     First  law  of  Arabic  notation  ?     Second  ? 


18  SIMPLE   NUMBERS. 

4th.   The  local  value  of  a  figure  depends  upon  its  place  from 

units  of  the  first  order,  not  upon  the  value  of  the  figures  at  the 

ri"-ht  of  it.     Thus,  in  425  and  400,  the  value  of  the  4  is  the 

same  in  both  numbers,  being  4  units  of  the  third  order,  or  4 

hundred. 

XoTE.  Care  should  be  taken  not  to  mistake  the  local  value  of  a 
figure  for  the  value  of  the  -whole  number.  For,  although  the  value  of 
the  4  (hundreds)  is  the  same  in  the  two  numbers,  425  and  400,  the  value 
of  the  whole  of  the  first  number  is  greater  than  that  of  the  second. 

0th.  Every  period  contains  three  figures,  (units,  tens,  and 
hundreds,)  except  the  left  hand  period,  which  sometimes  con- 
tains only  one  or  two  figures,  (units,  or  units  and  tens.) 

33.  As  we  liave  now  analyzed  all  the  principles  upon 
Avhich  the  writing  and  reading  of  whole  numbers  depend,  we 
will  present  these  principles  in  the  form  of  rules. 

RULE   FOR   NOTATION. 

I.  Beginning  at  the  left  hand,  write  the  figures  helonging  to 
the  liighest  period. 

II.  Write  the  hundreds,  tens,  and  units  of  each  successive 
period  in  their  order,  placing  a  cipher  wherever  an  order  of 
units  is  omitted. 

RULE   FOR   NUMERATION. 

I.  Separate  the  nnmhcr  into  periods  of  three  figures  each, 
commencing  at  the  right  hand. 

II.  Beginning  at  the  left  hand,  read  each  period  separately, 
and  give  the  name  to  each  period,  except  the  lost,  or  period 
of  units. 

34.  Until  the  pupil  can  write  numbers  readily,  it  may  be 
Avell  for  liim  to  write  several  periods  of  ciphers,  point  them  off, 
over  each  period  Avrite  its  name,  thus, 

Tnllinii.i,     r.illions,     IMillions,     Tliousnnds,     Units. 

000  ,    000,    000,     000  ,    000 


Fourth  principle  ?     "NATiat  raution  is  given  ?     Fifth  principle  ?     Rule 
for  notation  ?     Numeration  } 


NOTATION   AND   NUMERATION.  19 

and  then  Avrite  the  given  numbers  underneath,  in  their  appro- 
priate places. 

EXERCISES    IN    NOTATION    AND    NUMERATION. 

Express  the  following  numbei's  bj  figures  :  — 

1.  Four  hundred  thirty-six. 

2.  Seven  thousand  one  hundred  sixty-four. 

3.  Twenty-six  thousand  twenty-six. 

4.  Fourteen  thousand  two  hundred  eighty. 

5.  One  hundred  seventy-six  thousand. 

6.  Four  hundred  fifty  thousand  thirty-nine. 

7.  Ninety-five  million. 

8.  Four  hundred  thirty-three  million  eight  hundred  sixteen 
thousand  one  hundred  forty-nine. 

9.  Nine  hundred  thousand  ninety. 

10.  Ten  million  ten  thousand  ten  hundred  ten. 

11.  Sixty-one  billion  five  million. 

12.  Five  trillion  eighty  billion  nine  million  one. 

Point  ofl^,  numerate,  and  read  the  following  numbers:  — 


13. 

8240. 

17.             1010. 

21.              370005. 

14. 

400000. 

18.      57468139. 

22.      9400706342. 

15. 

308. 

19.             5628. 

23.          38429526. 

16. 

60720. 

20.   850026800. 

24.   74268113759. 

25. 

Write  seve 

n  million  thirty-six. 

26. 

Write  five 

hundred  sixty-three  th 

ousand  four. 

27. 

Write  one 

million  ninety-six  thou 

sand. 

28. 

Numerate  < 

^nd  read  9004082501. 

29. 

Numerate  j 

ind  read  25845039620 

47.                 * 

30.  A  certain  number  contains  3  units  of  the  seventh  order, 
6  of  the  fifth,  4  of  the  fourth,  1  of  the  third,  5  of  the  second, 
and  2  of  the  first ;  what  is  the  number  ? 

31.  What  orders  of  units  are  contained  in  the  number  29064S  ? 

32.  What  orders  of  units  are  contained  in  the  number 
1037050? 


20  SIMPLE   NUMBERS. 


ADDITION. 


MENTAL    EXERCISES. 

35.  1.  Ilcnry  gave  5  dollars  for  a  vest,  and  7  dollars  for 
a  coat ;  bow  much  did  be  pay  for  botb  ? 

Analysis.  He  gave  as  many  dollars  as  o  dollars  and  7  dollars, 
which  are  12  dollars.     Therefore  he  paid  12  dollars  for  both. 

2.  A  farmer  sold  a  pig  for  3  dollars,  and  a  calf  for  8  dol- 
lars ;  bow  much  did  be  receive  for  botb  ? 

3.  A  drover  bought  5  sheep  of  one  man,  9  of  another,  and 
3  of  another  ;  bow  many  did  be  buy  in  all  ? 

4.  How  many  are  2  and  G  ?  2  and  7  ?  2  and  9  ?  2  and  8  ? 
2  and  10  ? 

5.  IIow  many  are  4  and  5  ?  4  and  8  ?   4  and  7  ?  4  and  9  ? 

6.  How  many  are  6  and  4?   6  and  6  ?    6  and  9  ?   G  and  7  ? 

7.  How  many  are  7  and  7  ?  7  and  G  ?  7  and  8  ?  7  and  10  ? 
7  and  9  ? 

8.  How  many  are  5  and  4  and  6?  7  and  3  and  8  ?  G  and  9 
and  5  ? 

36.  From  the  preceding  operations  we  perceive  that 
Addition  is  the  process  of  uniting  several  numbers  of  the 

same  kind  into  one  equivalent  number. 

37.  The  Sum  or  Amount  is  the  result  obtained  by  the 
process  of  addition. 

38.  The  sign,  -|-,  is  called  plus,  "which  signifies  mo7-e. 
"When  placed  between  two  numbers,  it  denotes  that  they  are 
to  be  added ;  thus,  G  -f-  4,  shows  that  6  and  4  are  to  be  added. 

30.  The  sign,  =,  is  called  the  sign  of  equality.  "When 
placed  between  two  numbers,  or  sets  of  numbers,  it  signifies 
that  they  are  equal  to  each  other;  thus,  the  expression 
6  -|-  4  rr  10,  is  read  G  plus  4  is  equal  to  10,  and  denotes  that 
the  numbers  G  and  4,  taken  together,  equal  the  number  10. 

Define  addition.  The  sum  or  amount  ?  Sign  of  addition  ?  Of 
equality  ? 


ADDITION.  21 

CASE    I. 

40.  "When  the  amount  of  each  column  is  less 
than  10. 

1.  A  farmer  sold  some  hay  for  102  dollars,  six  cows  for 
1G2  dollars,  and  a  horse  for  125  dollars  ;  how  much  did  he  re- 
ceive for  all  ? 

OPERATION.  Analysis.     We  arrange  the  numhers  so 


to"- 


•a  .  ■ 


that  units  of  like  order  shall  stand  in  the 
III  same   column.     AVe  then  add  the  columns 

102  separately,  for   convenience   commencing  at 

1(32  the  right  hand,  and  write  each  result  under 

l^;-  the  column  added.     Thus,  we  have  5  and  2 

and  2  are  9,  the  sum  of  the  units  ;  2  and  6 

Amount,     389  are  8,  the  sum  of  the  tens  ;  1  and  1  and  1 

are  3,  the  sum  of  the  hundreds.     Hence,  ^he 

entire  amomit  is  3  hundi-ods  8  tens  and  9  units,  or  389,  the  Answer. 


(0.) 

dajs. 

437 
140 
321 


EXAMPLES    FOR    PRACTICE. 

(2.) 

(3.) 

(4.) 

pouuJs. 

rods. 

cents. 

132 

245 

312 

243 

321 

243 

324 

132 

412 

Ans.     699 

6.  What  is  the  sum  of  144,  321,  and  232  ?  A72S.    C97. 

7.  What  is  the  amount  of  122,  333,  and  401  ?     Ans.   856. 

8.  What  is  the  sum  of  42,  103,  321,  and  32  ?      Ans.    498. 

9.  A  drover  bought  three  droves  of  sheep.  The  first  con- 
tained 230,  the  second  425,  and  the  third  340 ;  how  many 
sheep  did  he  buy  in  all?  Ans.    995. 

CASE   II. 

41.  "When  the  amount  of  any  column  equals  or 
exceeds  10. 

1.   A  merchant  pays  725  dollars  a  year  for  the  rent  of  a 

Case  I  is  what  ?     Give  explanation.     Case  11  is  what  ? 


22 


SIMPLE    NUMBERS. 


Store,  475   dollars  for  a  clerk,  and  367  dollars  for  other  ex- 
penses ;  what  is  the  amount  of  his  expenses  ? 

Analysis.  Arranging  the  num- 
bers as  in  Case  I,  we  first  add  the 
column  of  units,  and  find  the  sum 
to  be  17  units,  which  is  1  ten  and 
7  units.  We  write  the  7  units  ni 
the  place  of  units,  and  the  1  ten  iu 
the  place  of  tens.  The  sum  of  the  • 
figures  in  the  column  of  tens  is  15 
tens,  which  is  1  hundred,  and  5 
tens.  We  write  the  5  tens  in  the 
place  of  tens,  and  the  1  hundred  in 
the  place  of  hundreds.  We  next 
add  the  column  of  hundreds,  and  find  the  sum  to  be  14  liundrcds, 
which  is  1  thousand  and  4  hundreds.  We  write  the  4  hundreds  in 
the  place  of  hundreds,  and  1  thousand  in  the  place  of  thousands. 
Lastly,  by  uniting  the  sum  of  the  units  with  the  sums  of  the  tens 
and  hundreds,  we  find  the  total  amount  to  be  1  thousand  o  hundreds 
6  tens  and  7  units,  or  1567. 

This  example  may  be  performed  by  another  method,  which 
is  the  common  one  in  practice.     Thus  : 


OPERATION. 

T3     .  at 

725 

475 

3G7 

Sum  of  the  units, 

17 

Sum  of  the  tens. 

15 

Sum  of  the  hundreds. 

14 

Total  amount, 

15G7 

OPEUATION 

725 
•475  •> 
3C)7 

15G7 


Analysis.  An-anging  tlie  numbers  as  before,  we 
add  the  first  column  and  find  the  sum  to  be  17  units; 
writing  the  7  units  under  the  column  of  units,-  we  add 
the  1  ten  to  the  column  of  tens,  and  find  the  sum  to  be 
10  tens  ;  writing  the  6  tens  under  the  column  tens,  we 
add  the  1  hundred  to  the  column  of  hundreds,  and  find 
the  sum  to  be  15  hundreds ;  as  this  is  the  last  column, 

we  write  down  its  amount,  15 ;  and  wc  have  the  lohole  amount,  15G7, 

as  before. 

Notes.  1.  T'nits  of  the  same  order  are  written  in  tlie  same  rolunm  ; 
and  when  the  sum  in  any  column  is  10  or  more  than  10,  it  procUiucs 
our  or  more  unitx  of  a  hi;^hfr  order,  whitli  must  be  addtd  to  the  next 
column.     This  process  is  sometimes  called  "  carryinn;  the  tens." 

2.  In  adding,  learn  to  pronounce  the  partial  results  without  namiii;? 
the  numbers  separately ;  thus,  instead  of  saying  7  and  5  are  12,  and 
6  are  17,  simply  pronounce  the  results,  7,  12,  17,  &c. 


Give  explanation. 
ing  the  tens  ? 


Second  explanation.     "SMiat  is  meant  by  carry- 


ADDITION.  23 

4L'^.  From  the  preceding  examples  and  illustrations  we 
deduce  the  following 

Rule.  I.  11  rite  the  numbers  to  he  added  so  tliat  all  the  units 
of  the  same  order  shall  stand  in  the  same  column  ;  that  is,  units 
under  units,  tens  under  tens,  8fc. 

II.  Commencing  at  units,  add  each  column  separateli/,  and 
write  tl/e  sum  underneath,  if  it  be  less  than  ten. 

III.  If'  the  sum  of  an?/  column  be  ten  or  more  than  ten,  write 
the  unit  figuret  only,  and  add  the  ten  or  tens  to  the  next  column. 

IV.  Write  the  entire  sum  of  the  last  column. 

Proof.  1st.  Begin  witli  the  right  hand  or  unit  column,  and 
add  the  figures  in  each  column  in  an  opposite  direction  from 
th:it  in  which  they  were  first  added  ;  if  the  two  results  agree, 
the  work  is  supposed  to  be  right.     Or, 

2d.  Separate  the  numbers  added  into  two  sets,  by  a  hori- 
zontal line ;  find  the  sum  of  each  set  separately ;  add  these 
sums,  and  if  tlie  amount  be  the  same  as  that  first  obtained,  the 
work  is  presumed  to  be  correct. 

Note.  By  the  methods  of  proof  here  given,  the  numbers  are  rmited 
in  new  combinations,  which  render  it  ahnost  impossible  for  two  pre- 
cisely simiUir  mistakes  to  occur. 

The  first  method  is  the  one  commonly  used  in  business. 

EXAMPLES    FOK   PRACTICE. 


(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

mill's. 

jncliL-3. 

tons. 

ft-et. 

buslii'ls. 

24 

321 

427 

1342 

3420 

48 

479 

321 

730G 

7021 

96 

1G5 

903 

5254 

327 

82 

327 

278 
1929 

8G29 
22531 

97 

250 

1292 

108G5 

Rule,  first  step  ?    Second  ?     Third  ?     Fourth  ?    Proof,  first  method  ? 
Second  ?     Upon  what  principle  are  these  methods  of  proof  founded  ? 


24  SIMPLE   KUMBERS. 


(7.) 

(8.) 

(9.) 

(10.) 

hours. 

years. 

gallons. 

rods. 

347 

7104 

3462 

47637 

506 

3762 

863 

3418 

218 

9325 

479 

703 

312 

4316 

84 

26471 

424 

2739 

57 

84 

11.  42  +  64 -|- 98 -f  70  +  37  ==  how  many  ?     Jns.  311. 

12.  312  -|-  425  +  107  -|-  391  +  76  =  Low  many  ? 

A71S.    1311. 

13.  1476 -j- 375  + 891 -f  66  4-80.=  how  many? 

Ans.   2888. 

14.  37042  -f  1379  -|-  809  +  127  +  40  rr:  how  many  ? 

Ans.    39397. 

15.  "What  is  the  sura  of  one  thousand  six  hundred  fifty-six, 
eight  hundred  nine,  three  hundred  ten,  and  ninety-four  ? 

Ans.    2869. 

16.  Add  forty-two  thousand  two  hundred  twenty,  ten  thou- 
sand one  hundred  five,  four  thousand  seventy-five,  and  five 
hundred  seven.  Ans.    56907. 

17.  Add  two  hundred  ten  thousand  four  hundred,  one  hun- 
dred thousand  five  liundred  ten,  ninety  thousand  six  hundred 
eleven,  forty-two  hundred  twenty-five,  and  eiglit  hundred 
ten.  Ans.    406556. 

18.  What  is  the  sum  of  the  following  nimihers :  seventy- 
five,  one  thousand  ninety-five,  six  thousand  four  hundred  thir- 
ty-five, two  hundred  sixty-seven  thousand,  one  thousand  four 
hundred  fifty-five,  twenty-seven  million  eighteen,  two  hundred 
seventy  million  twenty-seven  thousand?    Ans.    297303078. 

19.  A  man  on  a  journey  traveled  the  first  day  37  miles, 
the  second  33  miles,  the  third  40  miles,  and  the  fourth  35  miles  ; 
how  far  did  he  travel  in  the  four  days?  , 

20.  A  wine  merchant  has  in  one  cask  75  gallons,  in  another 
65,  in  a  third  57,  in  a  fourth  83,  in  a  fifth  74,  and  in  a  sixth 
67  ;  how  many  gallons  has  he  in  all?  Ans.    421. 


ADDITION.  25 

21.  An  estate  is  to  be  shared  equally  by  four  heirs,  and 
the  portion  to  each  heir  is  to  be  3754  dollars ;  what  is  the 
amount  of  the  estate?  ^ws.    15016  dollars. 

22.  IIow  many  men  in  an  army  consisting  of  52714  in- 
fimtry,  5110  cavalry,  6250  dragoons,  3927  light-horse,  928 
artillery,  250  sappers,  and  40(!  miners  ? 

23.  A  merchant  deposited  56  dollars  in  a  bank  on  Monday, 
74  on  Tuesday,  120  on  Wednesday,  96  on  Thursday,  170  on 
Friday,  and  50  on  Saturday ;  how  much  did  he  deposit  during 
the  week? 

24.  A  merchant  bought  at  public  sale  746  yards  of  broad- 
cloth, 650  yards  of  muslin,  2100  yards  of  flannel,  and  250 
yards  of  silk  ;  how  many  yards  in  all  ? 

25.  Five  persons  deposited  money  in  the  same  bank  ;  the 
first,  5897  dollars;  the  second,  12980  dollars;  the  third, 
65973  dollars  ;  the  fourth,  37345  dollars ;  and  the  fifih  as 
much  as  the  first  and  second  together ;  how  many  dollars  did 
they  all  deposit  ?  Ans.    141072  dollars. 

26.  A  man  willed  his  estate  to  his  wife,  two  sons,  and  four 
daughters  ;  to  his  daughters  he  gave  2630  dollars  apiece,  to 
his  sons,  each  4647  dollars,  and  to  his  wife  3595  dollars; 
how  much  was  his  estate  ?  Ans.    23409  dollars. 


(27.) 

(28.) 

(29.) 

(30.) 

(31.) 

476 

908 

126 

443 

180 

390 

371 

324 

298 

976 

915 

569 

503 

876 

209 

207 

245 

891 

569 

314 

841 

703 

736 

137 

563 

632 

421 

517 

910 

842 

234 

127 

143 

347 

175 

143 

354 

274 

256 

224 

536 

781 

531 

324 

135 

245 

436 

275 

463 

253 

RP. 


26  SIMPLE    NUMBEKS. 

32.  A  man  commenced  farming  at  the  west,  and  raised,  tlie 
first  year,  724  bushels  of  corn  ;  the  second  year,  'S-idS  bushels  ; 
the  third  year,  'JS72  bushels;  the  iburth  year,  9964:  bushels; 
the  lifth  year,  11078  bushels;  how  many  bushels  did  he  raise 
in  the  five  years  ?  Ans.   o5136  bushels. 

33.  A  has  oG48  dollars,  B  has  70oo  dollars,  C  has  429 
dollars  more  than  A  and  B  together,  and  D  has  as  many  dol- 
Lirs  as  all  the  rest ;  how  many  dollars  has  D  ?  How  many 
have  all?  Ans.  All  have  43590  dollars. 

34.  A  man  bought  three  houses  and  lots  lor  15780  dollars, 
and  sold  them  so  as  to  gain. 695  dollars  on  each  lot ;  for  how 
much  did  he  sell  them?  Ans.    17865  dollars. 

35.  At  the  battle  of  Waterloo,  which  took  place  June  18th, 
1815,  the  estimated  loss  of  the  French  was  40000  men ;  of 
the  .  Prussians,  38000  ;  of  the  Belgians,  8000  ;  of  the  Hano- 
verians, 3500  ;  and  of  the  English,  12000  ;  what  was  the  entire 
loss  of  life  in  this  battle  ? 

36.  The  expenditures  for  educational  purposes  in  New 
England  for  the  year  1850  were  as  follows :  Maine,  380623 
dollars;  New  Hampshire,  221146  dollars;  Vermont,  246604 
dollars  ;  Massachusetts,  1424873  dollars  ;  Rhode  Island, 
136729  dollars;  and  Connecticut,  430826  dollars;  what  was 
the  total  expenditure  ?  Ans.    2840801  dollars. 

37.  The  eastern  continent  contains  31000000  square 
miles;  the  western  continent,  13750000;  Australia,  Green- 
];md,  and  other  isl;uids,  5250000  ;  what  is  the  entire  area  of 
the  land  surface  of  the  glolie  ? 

38.  The  population  of  New  York,  in  1850,  was  515547; 
Boston,  136881;  riu'L-idelphia,  310015;  Chicago,  29963; 
St.  Louis,  77860  ;  New  Orleans,  116375;  Avhat  was  the  en- 
tiro  j)opulation  of  these  cities  ?  Ans.    1216671. 

39.  Tlu;  ])opulation  of  the  globe  is  es(iin;ited  ;i<  follows: 
North  America,  39257819;  South  Amerie:..  18373188;  Eu- 
rope, 265368216:  Asia,  630671661;  Africa,  61688779; 
Oceanic:!,  231II(IR2;  whnt  is  (he  total  jippulation  of  the 
globe  according  (o  this  estimate?  Ans.    1038803715. 


ADDITION. 


27 


40.  The  railroad  distance  from  Nbav  York  to  Albany  is  144 
miles  ;  fVoin  Albany  to  Jkitialo,  2'JS;  from  BuHalo  to  Cleveland, 
183  ;  from  Cleveland  to  Toledo,  109  ;  from  Toledo  to  Spring- 
field, oGo  ;  and  from  Springfield  to  St.  Louis,  95  miles;  wliat 
is  the  distance  from  Kew  York  to  St.  Louis? 

41.  A  man  owns  farms  valued  at  5G800  dollars;  city  lots 
valued  at  867G0  dollars;  a  house  worth  12500  dollars,  and 
other   property  to  the   amount  of  6785  dollars;  what  is  the 


entire  value  of  his 

property  ? 

Jns.    162845  dollars. 

(42.) 

(43.J 

(44.) 

(45.) 

15038 

2G881 

41919 

93803 

7404 

12173 

19577 

41371 

S4971 

39665 

74736 

110525 

30359 

33249 

66768 

102936 

G293 

6318 

12673 

17087 

2875 

4318 

7193 

13251 

IGGGO 

34705 

51365 

112110 

G4934 

80597 

155497 

220619 

8O0O1 

95299 

183134 

225255 

7444 

8624 

16845 

68940 

570G8 

53806 

111139 

176974 

17255 

18647 

35902 

86590 

32543 

41609 

82182 

149162 

40022 

35077 

75153 

109355 

5  GO  63 

46880 

132936 

283910 

338  GO 

41842 

82939 

112511 

17548 

26876 

44424 

72908 

28944 

36642 

65586 

157672 

16147 

29997 

52839 

86160 

38556 

44305 

83211 

119557 

234882 

262083 

522294 

839398 

39058 

39744" 

78861 

117787 

152526 

169220 

353428 

471842 

179122 

198568 

386214 

571778 

7626 

8735 

17005 

41735 

1218099 


1395860 


28  SIMPLE    NUMBERS. 


SUBTRACTION. 

MENTAL    EXERCISES. 

48.    1.    A  farmer,  having  l-i  cows,  sold  G  of  them;  how 
many  had  he  left? 

Analysis.     He  had  as  many  left  as  14  cows  less  6  cows,  wWch 
arc  8  coMS.     Therefore,  he  had  8  cows  left. 

2.  Stephen,  having  9  marbles,  lost  4  of  them ;  how  many 
had  he  left? 

3.  If  a  man  earn  10  dollars  a  week,  and  spend  6  dollars  for 
provision?,  how  many  dollars  has  he  left  ? 

4.  A  merchant,  having  IG  barrels  of  flour,  sells  9  of  them; 
how  many  has  he  left  ? 

5.  Charles  had  18  cents,  and  gave  10  of  them  for  a  book; 
how  many  had  he  left  ? 

G.   James  is  17   years  old,  and  his  sister  Julia,  is  5  years 
younger;  how  old  is  Julia? 

7.  A  grocer,  having  20  boxes  of  lemons,  sold  1 1   boxes ; 
how  many  boxes  had  he  left  ? 

8.  From  a  cistern  containing  25  barrels  of  water,  15  bai-- 
rels  leaked  out ;  how  many  barrels  remained  ? 

9.  Paid   IG  dollars  for  a  coat,  and  7  dollars   for  a  vest; 
how  much  more  did  tlie  coat  cost  than  the  vest  ? 

10.  How  many  arc  18  less  5  ?     17  less  8  ?     12  less  7  ? 

11.  How  many  are  20  less  14  ?     18  less  12?     19  less  11  ? 

12.  How  many  are  11  less  3?    IG  less  11?    19  less  8? 
20  less  9  ?    22  less  20  ? 

44.  Subtraction  is  the  process  of  determining  the  difler- 
CTice,  between  two  numbers  of  the  same  unit  value. 

4»"5.    The  Minuend  is  the  number  to  be  subtracted  from. 
46.    Tiie  Subtrahend  is  the  number  to  be  subtracted. 

T-^ — . 

Define  siibtrartion.     ^liimcnd.      Subtrahend. 


SUBTRACTION.  29 

4'^'*    The  Diiferenee  or  Eemaiuder  is  the  result  obtained 

b}'  the  process  ot'subti'action. 

Note.  The  iniiuiend  and  subtrahend  must  be  lilic  numbers ;  thus, 
5  dollars  from  9  dollars  Ivave  4  dollars  ;o  apples  from  9  apples  leave  4 
apples  ;  but  it  would  be  absurd  to  say  o  apples  fiom  9  dollars,  or  5 
dollars  from  9  apples. 

4:8.  The  sign,  — ,  is  called  minns,  which  signifies  less. 
When  placed  between  two  numbers,  it  denotes  that  the  one 
after  it  is  to  be  taken  from  the  one  before  it.  Thus,  8  —  6  =  2 
is  read  8  minus  G  equals  2,  and  denotes  that  G,  the  siihtrahend, 
taken  from  8,  the  minuend,  equals  2,  the  remainder. 

CASE    I. 

49.  When  no  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 

1.    From  574  take  323. 

OPERATION.  Analysis.     We  write  the  less  num- 

-Ti  ber  under  the   p-reater,  with  units  under 

o-,.^  units,  tens   under   tens,   ccc,   and    draw 

Subtrahena,         6:6  ^  j.^_^^  Underneath.     Then,  beginning  at 

Eemaindcr,  251  the  right  hand,  Ave   subtract    separately 

each  figure  of  the  subtrahend  from  the 
figure  above  it  in  the  minuend.  Thus,  3  from  4  leaves  1,  which  is  the 
difference  of  the  units  ;  2  from  7  leaves  .j,  the  difference  of  the  tens  ; 
3  from  5  leaves  2,  the  difference  of  the  hundreds.  Hence,  we  have 
for  the  whole  difference,  2  hundreds  5  tens  and  1  unit,  or  251. 

EXAMPLES   FOR   rRACTICE. 

(2.)  (3.)  (4.)  ■              (5.) 

5nnuend,             876  676  3C7  925 

SuUralH-nd,            334  415  152  213 

542  2G1  215  712 


Remainder, 


Case  lis  what?     Give  explanation. 


80  SIMPLE   NUMBERS. 

(6.)  (7.)  (8.)  /(9.) 

From  876  732  987  498 

Uake  523  522  782  178 

Remainders. 

10.  From  3276  take  2143.  1133. 

11.  From  7634  take  3132.  45U2. 

12.  From  41763  take  11521.  30242. 

13.  From  18346  take  5215.  13131. 

14.  From  397631  take  175321.  222310. 

15.  Subtract  47321  from  69524.  22203. 

16.  Subtract  16330  from  48673.  32343. 

17.  Subtract  291352  from  895752.  604400. 

18.  Subtract  84321  from  397502.  313241. 

19.  A  farmer  paid  645  dollars  for  a  span  of  horses  and 
a  carriage,  and  sold  them  for  522  dollars;  ho\v  much  did  he 
lose  ? 

20.  A  man  bought  a  mill  for  3724  dollars,  and  sold  it  for 
4856  dollars;  how  much  did  he  gain  .'*     Ans.    1132  dollars. 

21.  A  drover  bought  1566  sheep,  and  sold  435  of  them; 
how  many  had  he  left?  Ans.    1131  sheep. 

22.  A  piece  of  land  was  sold  for  2945  dollars,  which  was  832 
dollars  more  than  it  cost ;  what  did  it  cost  ? 

23.  A  gentleman  willed  to  his  son  15768  dollars,  and  to 
his  dniigliter  4537  dollars  ;  how  much  more  did  he  will  to  the 
son  than  to  the  daughter?  Ans.    11231  dollars. 

24.  A  merchant  sold  goods  to  the  amount  of  6742  dollars, 
and  by  so  doing  gained  2540  dollars ;  what  did  the  goods  cost 
him  ? 

25.  If  I  borrow  15  175  dollars  of  a  person,  and  pay  him 
4050  dollars,  how  much  do  I  still  owe  him  ? 

26.  In  1850  the  white  population  of  the  United  States  was 
19,553,068,  and  the  slave  population  3,201,313;  how  much 
was  the  difference? 

27.  Tiie  population  of  Great  Britain  in  1851  was  20,936,468, 
and  of  England  alone,  16,921,888  ;  what  was  the  difference? 


SUBTRACTION.  31 

\  CASE   II. 

50.  When  any  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 

1.    From  84G  take  351). 

OPERATION.  Analysis.     In  this   example  we 

(7)   (13)  (IC)  cannot  take  9  units  from  6  units. 

Minuend,            8     4  6  From  the  4  tens  we  take  1  ten,  which 

Subtrahend,         3      5  9  equals   10  units,   and  add  to  the  G 

~~,      [^^i      ^         units,  makiiif^  16  units  ;  9  units  from 
llemaiuder,  4      O       /  , ,.  .        ,  ,.  .  ,  •   , 

16    units  leave    /    units,   which  we 

write  in  the  remainder  in  units'  place.     As  we  have  taken  1  ten 

from  the  4  tens,  3  tens  only  are  left.     We  cannot  take  o  tens  from 

3  tens  ;  so  from  the  8  hundreds  we  take  1  hundred,  which  equals  10 

tens,  and  add  to  the  3  tens,  making  13  tens ;    5    tens   from    13 

tens  leave  8  tens,  which  we  write  in  the  remainder  in  tens'  place. 

As  we  have  taken  1  hundred  from  the  8  hundreds,  7  hundreds  only 

are  left ;  3  hundreds  from  7  hundreds  leave  4  hundreds,  which  we 

write  in  the  remainder  in  hundreds'  place,  and  we  have  the  whole 

remainder,  487.  , 

Note.  The  numbers  written  over  the  minuend  are  used  simply  to 
explain  more  clearly  the  method  of  subtracting ;  in  practice  the  pro- 
cess should  be  performed  mentally,  and  these  numbers  omitted. 

The  following  method  is   more   in  accordance  with   prac- 
tice. 

OPERATION.        Analysis.     Since  we  cannot  take  9  units  from  6 
1^:2  units,  we  add  10  units  to  6  units,  making  16  units; 

Jl  3  9  units  from  16  units  leave  7  units.     But  as  we  have 

846  added  10  units,  or  1  ten,  to  the  minuend,  we  shall 

359  have  a  remainder  1  ten  too  large,  to  avoid  which,  we 

.Qj  add  1  ten  to  the  5  tens  in  the  subtrahend,  making  6 

tens.  We  can  not  take  6  tens  from  4  tens  ;  so  we  add 
10  tens  to  4,  making  14  tens;  6  tens  from  14  tens 
leave  8  tens.  Now,  having  added  10  tens,  or  1  hundred,  to  the 
minuend,  we  shall  have  a  remainder  1  hundred  too  large,  unless  we 
add  1  hundred  to  the  3  hundreds  in  the  subtrahend,  making  4  hun- 
dreds ;  4  hundreds  from  8  hundreds  leave  4  hundreds,  and  we  have 
for  the  total  remainder,  487,  the  same  as  before. 


Case  II  is  what  ?     Give  explanation.     Second  explanation. 


32 


SIMPLE    NUMBERS. 


Note.  Tlie  process  of  adding  10  to  the  minuend  is  sometimes  called 
borrou'ing  10,  and  that  of  adding  1  to  the  next  figure  of  the  subtrahend, 
carrying  one. 

51.  From  the  preceding  examples  and  illustrations  we 
have  the  following  general 

Rule.  I.  Write  the  less  number  under  the  greater,  placing 
units  of  the  same  order  in  the  same  column. 

II.  Begin  at  the  right  hand,  and  take  each  figure  of  the  sub- 
trahend from  the  figure  above  it,  and  write  the  result  under- 
neath. 

III.  If  any  figure  in  the  subtrahend  be  greater  than  the  cor- 
responding figure  above  it,  add  10  /o  that  upper  figure  be- 
fore subtracting,  and  then  add  1  to  the  next  left  hand  figure  of 
the  subtrahend. 

Proof.  Add  the  remainder  to  the  subtrahend,  and  if  their 
sum  be  equal  to  the  minuend,  the  work  is  supposed  to  be  right. 

EXAMPLES    FOK    PRACTICE. 


(2.) 

(3.) 

(4.) 

(5.) 

Minuend, 

873 

7432 

1969 

814a 

Subtrahend, 

538 

C711 

1408 

4377 

Remainder, 

335 

(6.) 

(7.) 

(8.) 

(9.) 

gallons. 

bushels. 

miles. 

days. 

From 

3176 

9076 

7320 

5097 

Take 

2907 

4507 

3871 

3809 

(10.) 

(11.) 

(12.) 

(13.) 

dull.irs. 

rods. 

acivs. 

feet. 

From 

7G377 

67777 

900076 

767340 

Take 

457G1 

46699 

899934 

5039 

"What  do  Ave  mean  by  borrowing  10  ?    By  carrying  ?    Rule,  first  step  ? 
Second  ?     Third  ?     Proof  ? 


SUBTRACTIOX. 


83 


16. 
17. 
18. 
19. 
20. 
21. 
22. 

^2;). 

2L 
25. 

26. 


Ans. 

4111107. 

Ans. 

2479679. 

Ans 

.  935993. 

Ans 

.  608889. 

Ans. 

3968579. 

Ans.    50000001. 

Ans. 

3819851. 

Ans. 

,  800924. 

A^ns. 

7023024. 

14.  479  —  382  =z  how  many  ?  J?^s.    97. 

15.  6593  — 1807  =  how  many?  Ans.    4786. 
17380  —  3417  =  how  many  ?                  Ans.    13903. 
80014  —  43190  =z  liow  many  ?                Ans.    3GS24. 
282731  —  90756  =  how  many?           A71S.    191975. 
Flora  234100  take  9970.  . 
From  345673  take  124799. 
From  4367676  take  256569. 
From  3467310  take  987631. 
From  911000  take  5007. 
From  197()0:)0  take  1361111. 
From  290017  take  108045.     ' 
Take  3077097  from  7045676. 

27.  Take  9999999  from  60000000. 

28.  Take  220202  from  4040053. 

29.  Take  2199077  from  3000001. 

30.  Take  377776  from  8000800. 

31.  Take  501300347  from  1030810040. 

32.  Subtract  nineteen  thousand  nineteen  from  twenty  thou- 
sand ten.  Ans.    991. 

33.  From  one  miUion  nine  thousand  six  take  twenty  thou- 
sand four  hundred.  Ans.    988606. 

34.  What  is  the  difference  between  two  million  seven 
thousand  eighteen,  and  one  hundred  five  thousand  seven- 
teen ? 

EXAMPLES    COMDINING    AUDITION    AND    SUKTRACTIOX. 

t§Q.  1.  A  merchant  gave  his  note  for  5200  dollars.  He 
paid  at  one  time  2500  dollars,  and  at  another  175  dollars; 
what  remained  due  ?  Aiis..  2525  dollars. 

2.  A  traveler  who  was  1300  miles  from  home,  traveled 
homeward  235  miles  in  one  week,  in  the  next  275  miles,  in  the 
next  325  miles,  and  in  the  next  280  miles ;  how  far  had  he 
still  to  jro  before  he  would  reach  home?       Ans.    185  miles. 

3.  A  man  deposited  in  bank  8752  dollars ;  he  drew  out  at 
one  time  4234  dollars,  at  another  1700  dollars,  at  another  9G^ 

2* 


84  SIMPLE   NUMBERS. 

dollars,  and  at  another  49  dollars  ;  how  much  had  he  remain- 
ing in  bank  ?  Ans.    1807  dollars. 

4.  A  man  bought  a  farm  for  47 C")  dollars,  and  paid  750 
dollars  for  fencing  and  other  improvements  ;  he  then  sold  it  for 
384  dollars  less  than  it  cost  him  ;  how  much  did  he  receive 
for  it?  Ans.    5131  dollars. 

5.  A  forwarding  merchant  had  in  his  warehouse  7520  bar- 
rels of  flour;  he  shipped  at  one  time  1224  barrels,  at  another 
time  1500  barrels,  and  at  another  time  1805  barrels;  how 
many  barrels  remained? 

G.  A  had  450  sheep,  B  had  175  more  than  A,  and  C 
had  as  many  as  A  and  B  together  minus  114  ;  how  many 
sheep  had  C  ?  Ans.    9G1  sheep. 

7.  A  farmer  raised  1575  bushels  of  wheat,  and  900  bushels 
of  corn.  He  sold  807  bushels  of  wheat,  and  391  bushels  of 
corn  to  A,  and  the  remainder  to  B  ;  how  much  of  each  did 
he  sell  to  B  ?    Ans.    7G8  bushels  of  wheat,  and  509  of  corn. 

8.  A  man  traveled  G784  miles ;  2324  miles  by  railroad, 
1570  miles  in  a  stage  coach,  450  miles  on  horseback,  175 
miles  on  foot,  and  the  remainder  by  steamboat ;  how  many 
miles  did  he  travel  by  steamboat  ?  Ans.    22 G5  miles. 

9.  Three  persons  bought  a  hotel  valued  at  35G80  dollars. 
The  first  agreed  to  pay  7375  dollars:,  the  second  agreed  to 
pay  twice  as  much,  and  the  third  the  remainder ;  how  much 
was  the  third  to  pay  ?  Ans.    13555  dollars. 

10.  Borrowed  of  my  neighbor  at  one  time  750  dollars,  at 
another  time  379  dollars,  and  at  another  450  dollars.  Having 
paid  him  1000  dollars,  how  much  do  I  still  owe  him? 

Ans.    579  dollars. 

11.  A  man  wbrth  G709  d<illars,  received  a  legacy  of  3000 
dollars,  lie  spent  4379  dollars  in  traveling;  how  much  had 
he  left  ? 

12.  In  1850  the  number  of  white  males  in  (lie  United 
States  was  lOO^C.  102,  and  of  wliit(^  females  952(;0(;(;;  of 
these,  878G9G8  males,  and  8i)2o')Cu}  females  were  native 
\ion) ;  how  many  of  both  were  foreign  born  ?    Ans.    2240535. 


MULTIPLICATION,  86 


MULTIPLICATION. 

MENTAL    EXERCISES. 

53.  1.  What  will  4  pounds  of  sugar  cost,  at  8  cents  a 
pouud  ? 

Analysis.  Four  pounds  will  cost  as  much  as  the  price,  8  cents 
taken  4  times  ;  thus,  8-|-8-}-8-|-8=:32.  But  instead  of  adding, 
we  may  say,  —  since  one  pound  costs  8  cents,  4  pounds  will  cost  4 
times  8  cents,  or  '62  cents. 

2.  If  a  ream  of  paper  cost  3  dollars,  what  will  2  reams 
cost? 

3.  At  7  cents  a  quart,  what  will  4  quarts  of  cherries 
cost  ? 

4.  At  12  dollars  a  ton,  what  will  3  tons  of  hay  cost?  4 
tons  ?    5  tons  ? 

5.  There  are  7  days  in  1  week ;  how  many  days  in  6  weeks  ? 
in  8  wc(dvs  ? 

G.    What  will  9  chairs  cost,  at  10  shillings  apiece  ? 

7.  If  Henry  earn  12  dollars  in  1  month,  how  much  can  he 
earn  in  5  months?  in  7  months?  in  9  months? 

8.  What  will  11  dozen  of  eggs  cost,  at  9  cents  a  dozen?  at 
10  cents  ?  at  12  cents  ? 

9.  When  flour  is  7  dollars  a  barrel,  how  much  must  be 
paid  for  7  barrels  ?   for  9  barrels  ?   for  1 2  barrels  ? 

10.  At  9  dollars  a  week,  what  will  4  weeks'  board  cost  ? 
7  weeks'  ?   9  weeks'  ? 

11.  If  I  deposit  12  dollars  in  a  savings  bank  every  month, 
how  many  dollars  will  I  deposit  in  G  months  ?  in  8  months  ? 
in  9  months  ? 

12.  At  9  cents  a  foot,  what  will  4  feet  of  lead  pipe  cost? 
7  feet?    10  feet? 

13.  When  hay  is  8  dollars  a  ton,  how  much  will  3  tqns 
cost?   4  tons  ?   7  tons?   9  tons  ?    11  tons? 


86 


SIMPLE   NUMBERS. 


14.  "What  -will  be  the  cost  of  11  barrels  of  apples,  at  2  dol- 
lars a  barrel  ?   at  3  dollars  ? 

15.  At  10  cents  a  pound,  what  will  9  pounds  of  sugar  cost  ? 
11  pounds?    12  pounds? 

•54.  T5iiltiplication  is  the  process  of  taking  one  of  two 
given  numbers  as  many  times  as  there  are  units  in  the  other. 

c3S.    The  liiultiplicand  is  the  number  to  be  taken. 

c'°s$>.  The  Multiplier  is  the  number  which  shows  how 
many  times  the  multiplicand  is  to  be  taken. 

q37.  The  Product  is  the  result  obtained  by  the  process  of 
multiplication. 

58.    The  Factors  are  the  multiplicand  and  multiplier. 

Notes.  1.  Factors  are  producers,  and  the  multiplicand  and  mul- 
tiplier are  called  factors  because  they  produce  the  product. 

2.  ^Multiplication  is  a  short  method  of  performing  addition  when 
the  numbers  to  be  added  are  equal. 

«59.  The  &ign,'X,  plficed  between  two  numbers,  denotes 
that  they  are  to  be  multiplied  together ;  thus  9X6:=  54,  is 
read  9  times  G  equals  54. 

MTTLTIPLICATION    TABLE. 


1X1=1 

2X    1=    2 

3X    1=    3 

4X    1=    4 

IX    2=    2 

2X2=4 

3X  -2=    6 

4X2=8 

IX    3=3 

2  X    3=    6 

3X3=9 

4X    3  =  12 

IX    4=4 

2X    4=    8 

3X    4  =  12 

4  X    4  =  16 

1  X    o=    5 

2X    5  =  10 

3X    5  =  15 

4  X    5  =  20 

1X0=6 

2X    0=12 

3X    6  =  18 

4X    6  =  24 

*!  X    7=7 

2X    7  =  14 

3X    7  =  21 

4X    7  =  28 

1  X    8=    8 

2X    8  =  16 

3  X    8  =  21 

4X8  =  32 

1X9=9 

2X    9  =  18 

3  X    9  =  27 

4  X    9  =  3() 

1  X  10=10 

2  X  10  =  20 

3  X  10  =  30 

4  X  10  =  40 

1  X  11  =  11 

2  X  11  =  22 

3  X  1 1  =  33 

4  X  1 1  =  l-l 

1  X  12  =  12 

2  X  12  =  24 

3X  12  =  36 

4  X  12  =  48 

Define  multiplication.  Multiplicand.  ^Multiplier.  Prodtict.  Fac- 
tors.  Multiplication  is  a  short  niftb.od  of  what  ?  What  is  tiie  sign  of 
multiplication  ? 


MULTIPLICATION. 


87 


5X    1=    5 

GX     1=    6 

7X    1=    7 

8X    1=    8 

oX    2r=10 

GX    2  =  12 

7X    2=14 

8X    2=1G 

5X     3  =  1,; 

GX    3  =  18 

7X    3  =  21 

8  X    3  =  24 

5  X    4  =  20 

OX    4  =  24 

7X    4  =  28 

SX    4  =  32 

5  X    5  =  25 

GX    5  =  30 

7  X    5  =  35 

8  X    5  =  40 

0  X    C  =  30 

G  X    G  =  36 

7X    G  =  42 

8X    G  =  JS 

r>X     7  =  3J 

GX    7  =  42 

7X    7  =  49 

8  X    7  =  56 

oX    8  =  40 

0  X    8  =  48 

7  X    8  =  5G 

8  X    8  =  64 

oX    0  =  4.5 

6  X    0  =  54 

7  X    0  =  G3 

8X    9  =  72 

5  X  10  =  50 

6  X  10  =  GO 

7  X  10  =  70 

8  X  10  =  80 

5  X  11  =  55 

G  X  11  =  G6 

7X  11  =  77 

8  X  11=:  88 

5  X  12  =  60 

GX  12  =  72 

7  X  12  =  84 

8  X  12  =  96 

OX    1  = 

9 

10  X    1=    10 

IIX    1=    11 

12  X   1=    12 

OX    2  = 

18 

10  X   2=    20 

11  X   2=    22 

12  X   2=    24 

OX    3  = 

27 

10  X   3=    30 

11  X  3=    33 

12  X   3=    36 

OX    4  = 

36 

10  X  4=    40 

11  X  4=    44 

12  X  4=    48 

9X    5  = 

45 

10  X   5=    50 

11  X   o  —    00 

12  X   5=    60 

9X    6  = 

54 

10  X  6=    60 

11  X   6=    06 

12  X   6=    72 

9X    7  = 

63 

10  X   7=   70 

11  X   7=    77 

12  X   7=    84 

OX    8  = 

72 

10  X   8=    80 

11  X   8=    88 

12  X   8=    96 

9X    0  = 

81 

10  X   9=    90 

11  X  0=    99 

12  X   9=108 

9X  10  = 

90 

10X10  =  100 

11  xio  =  iio 

12X10=120 

9X11  = 

99 

10X11  =  110 

11  Xll  =  121 

12X11  =  132 

9X  12  = 

108 

10  X12  =  120 

11  X12  =  132 

12X12  =  144 

CASE    I. 

60.    When  tliG  multiplier  consists  of  one  figure. 

1.    Multiply  374  by  G. 

Analysis.  In  this  example  it  is» 
required  to  take  374  six  times.  If  we 
take  the  units  of  each  order  6  times, 
Ave  shall  take  the  entire  number  6 
times.  Therefore,  -writing  the  multi- 
plier under  the  unit  figure  of  the  mul- 
tiplicand, we  2)roceed  as  follows :  6 
times  4  units  are  24  units ;  6  times  7 
tens  are  42  tens  ;  6  times  3  hundreds 
are  18  hundreds ;  and  adding  these 
partial  products,  we  obtain  the  entire 
product,  2244. 


OPEKATION. 

Multiplicancl, 

374 

Multiplier, 

6 

units, 

24 

tens, 

42 

hundreds, 

18 

Product, 

2244 

Case  I  is  what  ?     Give  explanation. 


S8  SIMPLE   NUMBERS. 

The  operation  in  this  exainjile  may  be  performed  in  another 
way,  wliicli  is  the  one  in  common  use. 

OPERATION.       Analysis.     AVriting  the  numbers  as  before,  we 

o-i  begin  at  the  right  hand  or  unit  figure,  and  say:  6 

/»  times  4  units  are  24  units,  wliicli  is  2  tens  and  4 

units ;   write  the  4   units   in  the  product   in  units' 


2244  place,  and  reserve  the  2  tens  to  add  to  the  next  prod- 

uct ;  6  times  7  tens  are  42  tens,  and  the  two  tens  re- 
served in  the  last  product  added,  are  44  tens,  which  is  4  liaadicds 
and  4  tens ;  write  the  4  tens  in  the  product  in  tens'  place,  and  reserve 
the  4  hundreds  to  add  to  the  next  product ;  6  times  3  hundreds  are 
18  hundreds,  and  4  hundreds  added  are  22  hundreds,  which,  l^eing 
written  in  the  ])roduct  in  the  places  of  hundreds  and  thousands, 
gives,  for  the  entire  product,  2241. 

61.  The  unit  value  of  a  number  is  not  changed  by  re- 
peating the  number.  As  the  multiplier  always  expresses 
times,  the  product  must  have  the  same  unit  value  as  the  mul- 
tiplicand. But,  since  the  product  of  any  two  numbers  will  be 
the  same,  whichever  factor  is  taken  as  a  multiplier,  either 
factor  may  be  taken  for  the  multiplier  or  multiplicand. 

Note.  In  multiplying,  learn  to  pronounce  the  partial  results,  as  in 
addition,  without  naming  the  numbers  separately ;  thus,  in  the  last 
example,  instead  of  saying  6  times  4  are  24,  6  times  7  are  42  and  2  to 
carry  are  41,  6  times  3  arc  18  and  4  to  carry  are  22,  pronounce  only 
the  results,  21,  44,  22,  performing  the  operations  mentally.  This  will 
greatly  facilitate  the  process  of  multii)lymg. 

EXAMPLES    FOR    rUACTICE. 


(2.) 

(3.) 

(4.) 

MultipIicaiKl, 

7-324 

G812 

34G51 

•MultipliiT, 

4 
21)200 

0 

40872 

5 

rioiiiicf, 

17325} 

('>■) 

(G.) 

(7.) 

(8.) 

821oG 

92714 

28093 

-iG217 

7 

8 

9 

Second  explanation.     Repeating  a  number  has  what  effect  on  the 
unit  value  ?     The  product  must  be  of  the  same  kind  as  what  ? 


I 


MULTIPLICATION. 


89 


9.  Multi[)ly    327 W,  hy  o. 

10.  Multiply  840371  by  7. 

11.  Multii)ly  137G29  by  8. 

12.  Multiply    93702  by  3. 

13.  Multiply  543272  by  4. 

14.  Multiply  703164  by  9. 


Ans.  1G3730. 

Ans.  5882597. 

A/is.  1101032. 

Aus.  28128G. 

Alts.  21730S8. 

Ajis.  G32847G. 


15.  AVliut  will  be  the  cost  of  344  cords  of  wood  at  4  dol- 
lars a  cord  ?  Ans.    137G, 

16.  How  much  will  an  army  of  785G  men  receive  in  one 
week,  if  each  man  receive  6  dollars?       Ans.    4713G  dollars. 

17.  In  one  day  are  8G400  seconds;  how  many  seconds  in 
7  days  ?  Ans.  604800  seconds. 

18.  What  will  7 640  bushels  of  wheat  cost,  at  9  shillings  a 
bushel  ?  A}is.    G87G0  shillings. 

19.  At  5   dollars   an   acre,  what  Avill   2487  acres  of  land 
cost?  Ans.    12435  dollars. 

20.  In  one  mile  are  5280  feet;  how  many  feet  in  8  miles  ? 

Ans.    42240  feet. 


CASE    II. 


6sS.    "When  the  multiplier  consists  of  two  or  more 
fio-ures. 


1.   Muhiply  746  by  23. 

OPERATION. 
Multiplicand,  746 

Multiplier,  23 


2238 
1492 


Product, 


17158     23 


o  5  times  the  mul- 

^  ^  tiplicaiui. 
mS  times  the  mul- 
'   ( tiplicand. 

times  the  mul- 


tiplicand. 


Analysis.  AVrit- 
ing  the  multiplicand 
and  multiplier  as  in 
Case  I,  we  first  mul- 
ti])!)'  each  figure  in  the 
multiplicand  by  the 
unit  figure  of  the  mul- 
tiplier, precisely  as  in 
Case  I.  AVe  then  multijily  by  the  2  tens.  2  tens  times  6  units,  or  6 
times  2  tens,  are  12  tens,  equal  to  1  hundred,  and  2  tens  ;  we  place  the 
2  tens  under  the  tons  figure  in  the  product  already  obtained,  and  add 
the  1  hundred  to  the  next  hundreds  produced.  2  tens  times  4  tens 
are  8  hundreds,  and  the  1  hundred  of  the  last  product  added  are  9 
hundreds  ;  we  write  the  9  in  hundreds'  place  in  the  product.     2  tens 

Case  II  is  Avhat  ?     Give  explanation. 


40  SIMPLE    NUMBERS. 

times  7  hundreds  are  14  thousands,  equal  to  1  ten  thousand  and  4 
thousands,  wliich  we  -write  in  their  appropriate  places  in  the  ])roduct. 
Then  adding  the  two  products,  we  have  the  entire  product,  17158. 

Notes.  1.  When  the  multiplier  contains  two  or  more  figures,  the 
several  results  obtained  by  multiijlying  by  each  figure  are  called  7;a;//((/ 
products. 

•_'.  AN'hen  there  are  ciphers  hctwccn  the  significant  figures  of  the 
niuUipUer,  pass  over  them,  and  multiply  by  the  significant  figures  only, 

S3.  From  the  preceding  examples  and  illiistralions  we 
deduce  the  ibllowing  general 

Rule.  I.  Write  (he  multiplier  ^mder  the  7nulti2jlicand,pl(icinff 
units  of  the  same  order  under  each  otlier. 

II.  Multiply  the  multiplicand  by  each  fgure  of  the  multi- 
plier successively,  beginning  with  the  unit  fgure,  and  torite  the 
first  figure  of  each  partial  product  under  the  fgure  of  the  mul- 
tiplier used,  meriting  down  and  carrying  as  in  addition. 

III.  If  there  are  partial  products,  add  ihem,  and  their  sum 
icill  he  the  product  required. 

6 J:.  PnooF.  1.  IMultiidy  the  multiplier  by  (he  multipli- 
cand, and  if  the  product  is  the  same  as  the  first  result,  the 
Avork  is  correct.     Or, 

2.  Multiply  the  multiplicand  by  the  multiplier  diminished 
]>y  1,  and  to  the  product  add  the  multiplicand  ;  if  the  sum  he 
the  same  as  the  product  by  the  whole  of  the  multiplier,  the 
work  is  correct. 

EXAMPLES    FOR   rUACTXCE. 

(3.)  (4.) 

8721  17605 

47   '  '204 


(2.) 

Multiply 

4732 

liy 

36 

28392  61047  70420 

1410G  34884  35210 


Ans.        170352  409887  359M--0 


What  are  partial  products  ?  When  there  are  ciphers  in  the  multi- 
plier, how  proceed  ?  llule,  first  step  ?  Second  ?  Third  r  Proof, 
first  method  ?     Second  ? 


MULTIPLTCATION.  41 

(6.)  (7.) 

81092  379G7 

194  42G 


8.  IIow  many  yards  of  linen  in  759  pieces,  eacli  piece  con- 
taining 25  yards  ?  Ans.    18975  y<ards. 

9.  Sound  is  known  to  travel  about  1142  feet  in  a  second  of 
time  ;  how  far  will  it  travel  in  69  seconds  ? 

10  A  man  bought  36  city  lots,  at  475  dollars  each  ;  how 
mucli  did  they  all  cost  him?  Ans.    17100  dollars. 

11  What  would  be  the  value  of  867  shares  of  railroad 
stock,  at  97  dollars  a  share  ?  Ans.    8 1099  dollars. 

12.  How  many  pages  m  3475  book>;,  if  there  be  362 
pages  in  each  book  ?  Ans.    1257950  pages. 

13.  In  a  garrison  of  4507  men,  each  man  receives  annually 
208  dollars  ;  how  much  do  they  all  receive  ? 

14.  Multiply  7198  by  216.  Ans.    1554768. 

15.  Multiply  31416  by  175.  Ans.    5497800. 

16.  Multiply  7071  by  556.  Ans.    3931476- 

17.  Multiply  75649  by  579.  Ans.  43800771. 

18.  Multiply  15607  by  3094.  Ans.  48288058. 

19.  Multiply  79094451  by  76095.    Ans.    6018692248845. 

20.  Multiply  five  hundred  forty  thousand  six  hundred  nine^ 
by  seventeen  hundred  fifty.  Ans.    946065750. 

21.  Multiply  four  million  twenty-five  thousand  three  hun- 
dred ten,  by  seventy-five  thousand  forty-six. 

Ans.   302083414260. 

22.  Multiply  eight  hundred  seventy-seven  million  five  hun- 
dred ten  thousand  eight  hundred  sixty-four,  by  five  hundred 
forty-five  thousand  three  hundred  fifty-seven. 

A}is.   478556692258448. 

23.  If  one  mile  of  railroad  require  116  tons  of  iron,  Avorth 
65  dollars  a  ton,  what  will  be  the  cost  of  sufllcient  iron  to 
construct  a  road  128  miles  in  length?        Ans.  965120  dollars. 


42  SIMPLE   NUMBERS. 


CONTRACTIONS. 
CASK    I. 

G5.    When  the  multiplier  is  a  composite  number. 

A  Composite  Uumber  is  one  tli;it  may  be  j^ioduced  by 
multiplying  together  two  or  more  numbers;  thus,  18  is  a  com- 
posite number,  since  G  X  3  =  18  ;  or,  9  X  2  :=  18  ;  or,  3  X 
3X  2  —  18. 

06.  The  Component  Factors  of  a  number  are  the  sev- 
eral numbers  which,  multiplied  together,  produce  the  given 
number;  thus,  the  component  liictors  of  20  are  10  and  2, 
(10  X  2  =  20  ;)  or,  4  and  o,  (4  X  5  =  20  ,)  or,  2  and  2  and 
5,  (2  X  2  X  5  z=  20  ) 

XoiE.  The  pupil  must  not  confound  the  fnr/o)s  with  the  parts  of  a 
muiibtr.  'Ihus,  the  Jhc/ois  of  whuh  12  is  toniposed,  are  4  and  3, 
(^4  X  3=12;)  wlnle  the  par/s  of  which  12  is  composed  are  8  and  4, 
(8  -f-  4  =  12,)  or  10  fuid  2,  (10  +  2  =  12.)  The  factors  are  jnulliplicd, 
A\hde  the  pa)  ts  are  added,  to  produce  the  number. 

1.    "What  will  32  horses  cost,  at  174  dollars  apiece.'' 


MuUiplicanJ, 
1st  factor, 


OPKRATiox  Analysis.    The  fac- 

174  cost  of  1  horse.        tors  of  32   are  4  and 

A  8.     If  we  multiply  the 

cost   of  1  horse  by  4, 


G9G  cost  of  4  horses.      -^e  obtain  the  cost  of  4 

2d  factor,  8  horses ;  and  by  nudti- 

c  -/.o         X    r-o-Ti  plvinp:   the    cost    of  4 

rroiiuit,  0008  cost  or  62  horses.    ,'  •     "  ,     „  ,     . 

horses  by  8,  we  obtani 

the  cost  of  8  times  4  horses,  or  32  horses,  the  numLer  bought. 
G'7.    Hence  we  have  the  following 

Rile.  I.  Separate  the  composite  number  into  two  or  more 
fiictors. 

II.    Mtdtiplij  the  multiplicand  by  one  of  these  factors,  and 

What  arc  contractions?  Case  I  is  ^\hat?  Define  a  composite 
number.  Component  factors.  AVhat  caution  is  given  ?  Give  ex- 
phmation.      Kule,  first  step  ?     Second? 


MULTIPLICATION.  43 

that  product  ly  anotJier,  and  so  on  viitll  all  the  factors  have 
lean  used  succcssivelij ;  the  last  product  will  be  the  product  re- 
quired. 

Note.  The  product  of  any  number  of  factors  will  lie  the  ^aiiic  in 
Avhatcvcr  order  they  aic  niuhiphcd.  Tliu^;,  4  X  >^  X  5  =  60,  and 
5  X  4  X  3  =  G0. 

EXAMTLKS    FOR    TKACTICE. 

2.  Multiply  3472  by  48  r=  G  X  8.  Ans.    1CGG5G. 

3.  Multiply  147G1  by  64  =  8  X  8. 

4.  Multiply  87034  by  81  =  3  X  3  X  0.      Ans.    7049754. 

5.  Multiply  4732G  by  120  =:  G  X  5  X  4. 

G.    Multiply  G0315  by  9G.  Ans.    5700240. 

7.  Multii)ly  291042  by  125.  Ans.    3G380250. 

8.  If  a  vessel  sail  43G  miles  in  1  day,  bow  lar  Avill  slie  .-ail 
in  5G  (lays  ?  Ans.    2441G  miles. 

9.  IIow  much  will  72  acres  of  land  cost,  at  124  dollars  an 
acre  ?  Ans.    8928  dollars. 

10.  There  are  5280  feet  in  a  mile;  how  many  feet  in  84 
miles?  Ans.    443520  feet. 

11.  "What  will  120  yoke  of  cattle  cost,  at  125  dollars  a 
yok 


e? 


CASE    11. 

68.  When  the  multiplier  is  10,  100,  1000,  &c.- 
If  we  annex  a  cipher  to  the  multiplicand,  each  figure  is  re- 
moAed  one  place  toward  the  left,  and  consequently  the  value  of 
the  whole  number  is  increased  tenfold,  (Jl^.)  If  two  ciphers 
are  annexed,  each  figure  is  removed  two  places  toward  the 
left,  and  the  value  of  the  number  is  increased  one  hundred 
fold ;  and  every  additional  cipher  increases  the  value  teni'old. 
d.^.    Hence  the  following 

Rule.  Annex  as  many  cipliers  to  the  multiplicand  as  there 
are  cipliers  in  the  multiplier ;  the  number  so  formed  will  be 
the  product  required. 

Case  II  is  what  ?     Give  explanation.    Rule  ? 


44  SIMPLE   NUMBERS. 

EXAMPLES    FOR    PRACTICE. 

1.  Multiply  347  by  10.  Ans.    3470. 

2.  Multiply  4731  by  100.  Ans.    473100. 

3.  Multiply  13071  by  1000. 

4.  Multiply  89017  by  10000. 

5.  If  1  acre  of  land  cost  36  dollar?,  what   -will  10  acres 
cost?  Ans.    3 GO  dollars. 

G.   If  1  bushel  of  corn  cost  65  cents,  what  will  1000  bushels 
cost  ?  Ans.    G5000  cents. 

CASE    III. 

70,    When  there  are  ciphers  at  the  right  hand  of 
one  or  both  of  the  factors. 

1.   Multiply  1200  by  GO. 

OPERATIOX.  Analysis.     Both  multiplicand  and 

Multiplicand,  1200  multiplier  may  be  resolved  into  their 

jj  ,     ..                    /.Q  component  factors  ;  1200  into  12  and 

u  ip.er,  ^,^^^  ^^^^^  ^^  .^^^^  ^  ^^^^^  ^^      j^,  ^^^^^^ 

Product,  72000  several  factors  be  multiplied  together 

they  will  produce  the  same  product  as 
the  given  numbers,  (6Y.)  Thus,  12  X  6  ==72,  and  72  X  100  =r 
7200,  and  7200  X  10  ^  72JOO,  which  is  the  same  result  as  in  the 
operation.     Hence  the  followii; 


i^g 


Rule.  MuttijiJy  the' signijicant  Jigures  of  the  muItipUcnnd 
hj  tliose  of  the  multiplier,  and  to  tlie  product  annex  as  many 
ciphers  as  there  are  ciphers  on  the  right  of  both  factors. 


]Mult 
By 

iply 

EXAJIl'LES 

(2.)        ' 
4720 
340000 

1888 
141G    . 

1G04800000 

POU 

PRACTICE 

« 

.  I034(K)00 
10,)000 

5170 
1034 

108570000U000 

Case  III  is  what  ?     Give  explanation.     Rule. 


( 


MULTIPLICATION.  46 

4.  Multiply  70340  by  800400.        Ans.  563001 3G000. 

5.  Multiply  3400900  by  207000.     Ans.  70398G300000. 

6.  Multiply  634003000  by  40020.   Ans.  25372800060000. 

7.  Multiply  10203070  by  50302000. 

A71S.   513234827140000. 

8.  Multiply  30090800  by  600080.   Ans.   18056887264000. 

9.  Multi|)ly  eighty  million  seven  thousand  six  hundred,  by 
eight  million  seven  hundred  sixty.      Ans.    640121605776000. 

10.  Multiply  fifty  million  ten  thousand  seventy,  by  sixty- 
four  thousand.       ■  Ans.    3200044480000. 

11.  Multiply  ten  million  three  hundred  fifty  thousand  one 
hundred,  by  eighty  thousand  nine  hundred. 

Ans.   837323090000. 

12.  There  are  296  members  of  Congress,  and  each  one  re- 
ceives a  salary  of  3000  dollars  a  year ;  how  much  do  they  all 
receive  ? 

EXAMPLES    COMBINING    ADDITION,    SUBTRACTION,    AND 
MULTIPLICATION. 

1.  Bought  45  cords  of  wood  at  4  dollars  a  cord,  and  9  loads 
of  hay  at  13  dollars  a  load;  what  was  the  cost  of  the  wood 
and  hay  ?  Ans.    297  dollars. 

2.  A  merchant  bought  6  hogsheads  of  sugar  at  31  dollars 
a  hogshead,  and  sold  it  for  39  dollars  a  hogshead ;  how  much 
did  he  gain? 

3.  Bought  288  barrels  of  flour  for  1875  dollars,  and  sold 
the  same  for  9  dollars  a  barrel ;  how  much  was  the  gain  ? 

Ans.    717  dollars. 

4.  If  a  young  man  receive  500  dollars  a  year  salary  and 
pay  240  dollars  for  board,  125  dollars  for  clothing,  75  dollars 
for  books,  and  50  dollars  for  other  expenses,  how  much  will 
he  have  left  at  the  end  of  the  year  ?  Ans.    10  dollars. 

5.  A  farmer  sold  184  bushels  of  wheat  at  2  dollars  a 
bushel,  for  which  he  received  67  yards  of  cloth  at  4  dollars  a 
yard,  and  the  balance  in  groceries ;  how  much  did  his  gro- 
ceries cost  him  ? 


46  SIMPLE   NUMBERS. 

6.  A  sold  a  farm  of  320  acres  at  36  dollars  an  acre  ;  B 
sold  one  of  244  acres  at  48  dollars  an  acre ;  uliicli  received 
the  greater  sum,  and  how  much?  Ans.    B,  102  dollars. 

7.  Two  persons  start  from  the  same  point  and  travel  in 
opposite  directions,  one  at  the  rate  of  35  miles  a  day,  and  ihe 
other  29  miles  a  day  ;  bow  far  apart  will  they  be  in  16  days? 

Ans.    1024  miles. 

8.  A  merchant  tailor  bought  14  bales  of  cloth,  each  bale 
containing  26  pieces,  and  each  piece  43  yards;  how  many 
yards  of  cloth  did  he  buy?  Ans.    15652  yards. 

9.  If  a  man  have  an  income  of  3700  dollars  a  year,  and  his 
daily  expenses  be  4  dollars ;  what  will  he  save  in  a  year,  or 
365  days  ?  Ans.    2240  dollars. 

10.  A  man  sold  three  houses  ;  for  the  first  he  received 
2475  dollars,  for  the  second  840  dollars  less  than  he  received 
for  the  first,  and  for  the  third  as  much  as  for  the  other  two; 
how  much  did  he  receive  for  the  three  ?     Ans.    8220  dollars. 

11.  A  man  sets  out  to  travel  from  Albany  to  Buffalo,  a 
distance  of  336  miles,  and  walks  28  miles  a  day  for  10  days ; 
how  far  is  he  from  BuflTalo? 

12.  Mr.  C  bought  14  cows  at  23  dollars  each,  7  horses  at 
96  dollars  each,  34  oxen  at  57  dollars  each,  and  300  sheep  at 
2  dollars  each ;  he  sold  the  whole  for  3842  dollars  ;  how 
much  did  he  gain?  Ans.    310  dollars. 

13.  A  drover  bought  164  head  of  cattle  at  36  dollars  a 
head,  and  850  sheep  at  3  dollars  a  head  ;  how  much  did  he 
pay  for  all  ? 

14.  A  banker  has  an  income  of  14760  dollars  a  year;  he 
pays  1575  dollars  for  house  rent,  and  four  times  as  much  for 
fitmily  expenses  ;  how  much  does  he  save  annually  ? 

Ans.    6885  dollars. 

15.  A  flour  merchant  bought  936  barrels  of  flour  at  9  dol- 
lars a  barrel;  he  sold  480  ban-els  at  10  dollars  a  barrel,  and 
the  remainder  at  8  dollars  a  barrel ;  how  much  did  he  gain  or 
lose  ?  Ans.    Gained  24  dollars. 


DIVISION.  47 


DIVISION. 


MENTAL    EXERCISES. 


71 .  1.  IIow  many  hats,  at  4  dollars  apiece,  can  be  bought 
for  20  dollars  ? 

Analysis.  Since  4  dollars  will  buy  one  hat,  20  dollars  will  buy 
as  many  hats  as  4  is  contained  times  in  20,  which  is  o  times.  There- 
fore, 5  hats,  at  4  dollars  apiece,  can  be  bouglit  for  20  dollars. 

2.  A  man  gave  16  dollars  for  8  barrels  of  apples;  what 
was  the  cost  of  each  barrel  ? 

3.  If  1  cord  of  wood  cost  3  dollars,  how  many  cords  can 
be  bought  for  15  dollars  ? 

4.  At  6  shillings  a  bushel,  how  many  bushels  of  corn  can 
be  bought  for  24  shillings  ? 

5.  When  flour  is  6  dollars  a  barrel,  how  many  barrels  can 
be  bought  for  30  dollai-s  ? 

6.  If  a  man  can  dig  7  rods  of  ditch  in  a  day,  how  many 
davs  will  it  take  him  to  dig  28  I'ods  ? 

7.  If  an  oi-chard  contain  56  trees,  and  7  trees  in  a  row, 
how  many  rows  are  there  ? 

8.  Bought  6  barrels  of  flour  for  42  dollars  ;  what  was  the 
cost  of  1  barrel  ? 

9.  If  a  farmer  divide  21  bushels  of  potatoes  equally 
among  7  laborers,  how  many  bushels  will  each  receive  ? 

10.  How  many  oranges  can  be  bought  for  27  cents,  at  3 
cents  each  ? 

11.  A  farmer  paid  35  dollars  for  sheep,  at  5  dollars  apiece ; 
how  many  did  he  buy  ? 

12.  IIow  many  times  4  in  28  ?  in  IG  ?  in  36  ? 

13.  How  many  times  8  in  40  ?  in  56  ?  in  64  ? 

14.  How  many  times  9  in  36?  in  63?  in  81  ? 

15.  IIow  many  times  7  in  49  ?  in  70  ?  in  84  ? 


48  SIMPLE   NUMBERS. 

7?2.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another. 

73.  The  Dividend  is  the  number  to  be  divided. 

74.  The  Divisor  is  the  number  to  divide  by. 

7^1,  The  Quotient  is  the  result  obtained  by  the  process  of 
division,  and  shows  how  many  times  the  divisor  is  contained 
in  the  dividend. 

Notes.  1.  "When  the  dividend  does  not  contain  the  divisor  an  exact 
number  of  times,  the  part  of  the  di\idend  left  is  called  the  remainder, 
and  it  must  be  less  than  the  divisor. 

2.  As  the  remainder  is  always  a  part  of  the  diA-idend,  it  is  always 
of  the  same  name  or  liind. 

3.  "When  there  is  no  remainder  the  division  is  said  to  be  complete. 

76,  The  sign,  -4-,  placed  between  two  numbers,  denotes 
division,  and  shows  that  the  number  on  the  left  is  to  be  divided 
by  the  number  on  the  right.  Thus,  20  -7-  4  :=:  5,  is  read,  20 
divided  by  4  is  equal  to  5. 

Division  is  also  indicated  by  writing  the  dividend  ahovcy  and 

.  .  12 

the  divisor  heloxo  a  short  horizontal  line ;  thus,  —  =  4,  shows 

that  12  divided  by  3  equals  4. 

CASE   I. 

77.  When  the  divisor  consists  of  one  figure. 

1.    How  many  times  is  4  contained  in  848? 

OPERATION.  Analysis.     After  writing  the  divisor 

on  the  left  of  the  dividend,  with  a  Hne 

,  N.  o  i  q'  between  them,  we  begin  at  the  left  hand 

^ and  say :  4  is  contained  in  8  hundreds, 

Qnotient,  212  2  hundreds  times,   and  write  2  in  hun- 

dreds' place  in  the  quotient ;  then  4  is 
contained  in  4  tens  1  ten  times,  and  write  the  1  in  tens'  place  in  the 
quotient ;  then  4  is  contained  in  8  units  2  units  times  ;  and  writing  the 
2  in  units'  place  in  the  quotient,  we  have  the  entire  quotient,  212. 

Define  division.  Dividend.  Divisor.  Quotient.  Kcmnindcr. 
"What  is  complete  division  ?  "Wliat  is  the  sign  of  division.  Caso  I  is 
what?     Give  first  exi)lntiation. 


DIVISION.  4$ 

2.  How  many  times  is  4  contained  in  2884  ? 
OPERATION.        Analysis.     As  we  cannot  divide  2  thousands  by 
4V^884  ^'  ^^'^  ^'^^^^  ^^^  ^  thousands  and  the  8  hundreds  to- 

^^ gether,  and  say,  4  is  contained  in  28  hundreds  7  hun- 

721  dreds  times,  which  we  write  in  hundreds'  place  in 

the  quotient ;  then  4  is  contained  in  8  tens  2  tens 
times,  wliich  we  write  in  tens'  place  in  the  quotient ;  and  4  is  con- 
tained in  4  units  1  unit  time,  which  we  write  in  units'  place  in  tlie 
quotient,  and  we  have  the  entire  quotient,  721. 

3.  How  many  times  is  6  contained  in  1824  ? 

OPERATION.        Analysis.     Beginning  as  in  the  last  example,  we 
6)  1824  say,  6  is  contained  in  18  hundreds  3  hundi-eds  times, 

which  we  write  in  hundreds'  place  in  the  quotient ; 

^^^  then  6  is  contained  in  2  tens  no  times,  and  we  write 

a  cipher  in  tens'  place  in  the  quotient;  and  taking  the  2  tens  and  4 
units  together,  6  is  contained  in  24  units  4  units  times,  which  we 
write  in  units'  place  in  the  quotient,  and  we  have  304  for  the  entire 
quotient. 

4.  How  many  times  is  4  contained  in  943  ? 

OPERATION.    ■•  Analysis.     Here  4  is  contained  in  9 

1  \n_)  3  hundreds  2  hundreds  times,  and  1  hundred 

over,  which,  united  to  the  4  tens,  makes 

235  ...  3  Rem.  14  tens ;  4  in  14  tens,  3  tens  times  and  2 
tens  over,  which,  united  to  the  3  units, 
make  23  units ;  4  in  23  units  5  units  times  and  3  units  over.  The 
3  which  is  left  after  performing  the  division,  should  be  divided  by  4  ; 
but  the  method  of  doing  it  cannot  be  explained  until  we  reach 
Fractions ;  so  we  merely  indicate  the  division  by  placing  the  divisor 
under  the  dividend,  thus,  |.  The  entire  quotient  is  written  235|, 
wliich  may  be  read,  two  hundred  thirty-five  and  three  divided  by 
four,  or,  two  hundred  thirty-five  and  a  remainder  of  three. 

From  the  foregoing  examples  and  illustrations,  we  deduce 
the  following 


*o 


Rule.    I.    Write  the  divisor  at  the  left  of  the  dividend,  with 
a  line  between  them. 

• , ^*— —  ^y 

Second.     Third.     Rule,  first  step? 
RP  8 


50 


SIMPLE    NUMBERS. 


II.  Beginning  at  the  left  hand,  divide  each  figure  of  the 
dividend  by  the  divisor,  and  write  the  result  under  the  divi- 
dend. 

HI.  If  there  be  a  remainder  after  dividing  any  figure,  re- 
gard it  as  prefixed  to  the  figure  of  the  next  lower  order  in  tlie 
dividend,  and  divide  as  before. 

IV.  Should  any  figure  or  part  of  the  dividend  be  less  than 
the  divisor,  write  a  cipher  in  the  quotient,  and  prefix  the  num- 
ber to  the  figure  of  the  next  lower  order  in  the  dividend,  and 
divide  as  before. 

Y.  Jf  titer e  be  a  remainder  after  dividing  the  last  figure, 
place  it  over  the  divisor  at  the  rigid  hand  of  the  quotient. 

Proof.  Multiply  the  divisor  and  quotient  together,  and  to 
the  product  add  the  remainder,  if  any;  if  the  result  be  equal 
to  the  dividend,  the  work  is  correct. 

Notes.  1.  This  method  of  proof  depends  on  the  fact  that  division  is 
the  reverse  of  multiplication.  The  dividend  answers  to  the  product,  the 
divisor  to  one  oi  the  factors,  and  the  quotient  to  the  other, 

2.  In  multiplication  the  two  factors  are  giA-en,  to  find  the  product  . 
in  di\T.sion,  the  product  and  one  of  the  factors  are  given  to  find  the 
other  factor. 


EXAMPLES 

FOR   PRACTICE. 

1.    Divide  7824  by  6. 

OPERATION. 
Divisor.       6)7824       Dividend. 

1304       Quotient. 

PROOF. 
1304       Quotient. 
6       Divisor. 

7824       Dividend. 

(2.)                                       (3.) 
4)05432                               5)89135 

6)1789^ 

7)4708935 


(G.) 
8)1462376 


9)7468542 


Second  step?     Third?     Fourth?     Fifth?     Proof?    ITow  does  divis- 
ion differ  from  multiplication  ? 


DIVISION. 


51 


8.  Divide  3102455  by  5. 

9.  Divide  1762891  by  4. 

10.  Divide  546215747  by  11. 

11.  Divide  30179G24by-12. 

12.  Divide  9254671  by  9. 


Quotients. 


Quotients. 
G20491. 
440722a. 
49655977. 
25 149 68  jC 
1028296^." 
Rem. 


13.  Divide  7341568  by  7. 

14.  Divide  3179632  by  5. 

15.  Divide  19038716  by  8. 

16.  Divide  84201763  by  9. 

17.  Divide  2947691  by  12. 

18.  Divide  42084796  by  6. 

Sums  of  quotients  and  remainders,    20680083.  28. 

19.  Divide  47645  dollars  equally  among  5  men;  how 
mucli  -will  each  receive  ?  Ans.^  9529  dollars. 

20.  In  one  week  are  7  days;  how  many  weeks  in  17675 
days?  Ans.    2525  weeks. 

21.  How  many  barrels  of  flour,  at  6  dollars  a  barrel,  can  be 
bought  for  6756  dollars?  Ans.    1126  barrels. 

22.  Twelve  things  make  a  dozen ;  how  many  dozen  in 
46216464?  Ans.   3851372  dozen. 

23.  How  many  barrels  of  flour  can  be  made  from  347560 
bushels  of  wheat,  if  it  take  5  bushels  to  make  one  barrel  ? 

Ans.    69512  barrels. 

24.  If  there  be  3240622  acres  of  land  in  11  townships, 
how  many  acres  in  each  tOAvnship  ? 

25.  A  gentleman  left  his  estate,  worth  38470  dollars,  to  be 
shared  equally  by  his  wife  and  4  children  ;  how  much  did 
each  receive  ?  Ans.   7694  dollars. 


CASE    II. 

78.   When  the  divisor  consists  of  two  or  more  figures. 

Note.    To  illustrate  more  clearly  the  method  of  operation,  we  will 
first  take  an  example  usually  performed  by  Short  Division. 


Case  n  is  what  ? 


52  SIMPLE    NUMBERS. 

1.  How  many  times  is  8  contained  in  2528  ? 
OPERATION.  Analysis.     As  8  is  not  contained  in  2  thou- 

8  "i  25'^8  (  316         sands,  we  take  2  and  5  as  one  number,  and  con- 
ni  sider  how  many  times  8  is  contained  in  this 

partial  dividend,  25  hundreds,  and  find  that  it 

12  is  contained  '3  hundreds  times,  and  a  remainder, 

8  To  find  tliis  remainder,  we  muhiply  the  divisor, 

"TT  8,  by  the  quotient  figure,  3  hundreds,  and  sub- 

tract the  product,  24  hundreds,  from  the  par- 
**'"  tial  dividend,  25  hundreds,  and  there  remains 

1  hundred.  To  tliis  remainder  we  bring  down 
the  2  tens  of  the  dividend,  and  consider  the  12  tens  a  second  partial 
dividend.  Then,  8  is  contained  in  12  tens  1  ten  time  and  a  remain- 
der ;  8  multiplied  by  1  ten  produces  8  tens,  which,  subtracted  from 
12  tens,  leave  4  tens.  To  this  remainder  we  bring  down  the  8  units, 
and  consider  the  48  units  the  third  partial  dividend.  Tlien,8  is  con- 
tained in  48  units  6  units  times.  Multiplying  and  subtracting  as 
before,  we  find  that  nothing  remains,  and  we  have  for  the  entii'e 
quotient,  316. 

2.  How  many  times  is  23  contained  in  4807  ? 

OPERATION.  Analysis.     AVe  first  find  how 

Divisor.  Divid'd.  Quotient.  many  times  23  is  contained  in  48, 

23  )  4807  (  209  the  first  partial  dividend,  and  place 

46  the  result  in  the   quotient  on  the 

~T~~I  right  of  the   dividend.     "We  then 

niultiijly   the   divisor,    23,    bv   the 
907  • 

"    *  quotient  figure,  2,  and  subtract  the 

product,  46,  from  the  part  of  th^ 
dividend  used,  and  to  the  remainder  bring  down  the  next  figure  of 
the  dividend,  which  is  0,  making  20,  for  the  second  partial  dividend. 
Then,  since  23  is  contained  in  20  no  times,  we  place  a  cl|)hcr  in  the 
quotient,  and  bring  down  the  next  figure  of  the  dividend,  making  a 
third  partial  dividend,  207  ;  23  is  contained  in  207,  9  times ;  multi- 
plying and  sul)tracting  as  before,  notliing  remains,  and  we  have  for 
the  entire  quotient,  209. 

Notes.  1.  AMicn  the  process  of  dividintj  is  performed  mentally,  and 
the  results  only  are  written,  as  in  Case  I,  the  operation  is  termed  Short 
Division, 

2.  When  the  -whole  process  of  division  is  written,  the  operation  is 
termed  Loncf  Division, 

Give  first  explanation.  Second.  "WTiat  is  long  division  ?  "SMiat  is 
short  division  ■     AMien  is  each  used  ? 


DIVISION.  53 

3.  Short  Division  is  generally  used  ■vvhcn  the  divisor  is  a  number 
that  -will  allow  the  process  of  dividing  to  be  performed  mentally. 

From  the  preceding  illustrations  we  derive  the  following 
general 

Rule.  I.  Write  the  divisor  at  the  left  of  the  dividend,  as 
in  short  division. 

II.  Divide  the  least  number  of  the  left  hand  fgures  in  the 
dividend  that  icill  contain  the  divisor  one  or  more  times,  and 
place  the  quotient  at  the  right  of  the  dividend,  with  a  line  be- 
tween them. 

III.  Multiply  the  divisor  by  this  quotient  fgure,  suhtract 
the  product  from  the  partial  dividend  used,  and  to  the  remain- 
der bring  down  the  next  figure  of  the  dividend. 

IV.  Divide  as  before,  until  all  the  figures  of  the  dividend 
have  been  brought  down  and  divided. 

V.  If  any  partial  dividend  will  not  contain  the  divisor, 
place  a  cipher  in  the  quotient,  and  bring  down  the  next  figure 
of  the  dividend,  and  divide  as  before. 

VI.  If  there  be  a  remainder  after  dividing  all  the  figures  of 
the  dividend,  it  must  be  written  in  the  quotient,  with  the  divi- 
sor underneath. 

XoTES.  1.  If  any  remainder  be  equal  to,  or  greater  than  the  divisor, 
the  quotient  figure  is  too  small,  and  must  be  increased. 

2.  If  the  product  of  the  di^^sor  by  the  quotient  figure  be  greater 
than  the  partial  dividend,  the  quotient  figure  is  too  large,  and  must  be 
diminished. 

7f?.    Proof.     1.  The  same  as  in  short  division.     Or, 

2.  Subtract  the  remainder,  if  any,  from  the  dividend,  and 
divide  the  difference  by  the  quotient ;  if  the  result  be  the  same 
as  the  given  divisor,  the  work  is  correct. 

80.  The  operations  in  long  division  consist  of  five  prin- 
cipal steps,  viz. :  — 

1st.     "Write  down  the  numbers. 

Rule,  first  step  ?  Second  ?  Third  ?  Fourth  ?  Fifth  ?  Sixth  ?  Fkst 
dii'ection  ?     Second  ?     Proof  ?     Recapitulate  the  steps  in  their  order. 


54: 


SIMPLE    NUMBERS. 


2d.  Find  Iioav  many  times. 

3d.  Multiply. 

4tli.  Subtract. 

5  ill.  liriiig  down  another  figure. 


EXAMPLES    FOR    PRACTICE. 

3.    Find  how  many  times  36  is  contained  in  11798. 

OPERATIOX.  PROOF  BY  MULTIPLICATION. 


Divisor. 

Diviileuil. 

36)  11798 
108 

99 
72 

( 

327 

Quotient. 

327 
36 

1962.. 
981 

Quotient. 
Divisor. 

278 
252 

26 

Eemaintler. 

11772 
26 

Remainder. 

11798 

Dividend. 

4.    Find  how  many  times  82  is  contained  in  89634. 


OPERATION. 

82  )  80634  (  1093 
82 


PROOF   BY   DIVISION. 
89634  Dividend. 

8  Remainder. 


763 

738 


254 
246 

8 


Quotient.        1093  )  89626  (  82        Divisor. 

8744 

2186 
2186 


5.    Find  how  many  limes  154  is  contained  in  32740. 

Ans.  958. 

Aiis.  7198. 

Ans.  31416. 

Ans.  7071. 

A71S.  83209. 


6.  Divide  32572  by  34. 

7.  Divide  1554768  by  216. 

8.  Divide  5497800  by  175. 

9.  Divide  393M76  by  550. 
10.  Divide  10983588  by  132. 


DIVISION, 


55 


11. 

Divide 

12. 

Divide 

13. 

Divide 

14. 

Divide 

1.1 

Divide 

16. 

Divide 

17. 

Divide 

18. 

Divide 

19. 

Divide 

20. 

Divide 

21. 

Divide 

73484248  by  19. 
8121918  by  21. 
10.557312  by  16. 
93S40  by  63. 
352417  by  29. 
51«46734  by  102. 
1457924G51  by  1204. 
729386  by  731. 
4843167  by  3605. 
49816657  by  9i01. 
75867308  by  10115. 


Ans.  3867592. 
Ans.  386758. 
Ans.      659832. 


Divide  28101418481  by  1107. 
Divide  65358547823  by  2789. 
Divide  102030405060  by  123456. 
Divide  48659910  by  54001. 
Divide  233183S961  by  6739549. 


Hem. 

33. 

Rem. 

9. 

Rem. 

32. 

Rem. 

1051. 

Rem. 

579. 

Rem. 

1652. 

Rem. 

6884. 

Rem. 

4808. 

Quotients. 

Rem. 

25385201. 

974. 

23434402. 

645. 

826451. 

70404. 

901. 

5009. 

346. 

7. 

22. 

23. 
24. 
25. 
26. 

27.  A  railroad  cost  one  million  eight  hundred  fifty  thousand 
four  hundred  dolhirs,  and  was  divided  into  eighteen  thousand 
five  hundred  and  four  shares ;  what  was  the  value  of  eaeh 
share?  A.ns.    100  dollars. 

28.  If  a  tax  of  sBventy-two  million  three  hundred  twenty 
thousand  sixty  dollars  be  equally  assessed  on  ten  thousand 
seven  hundred  thirty-five  towns,  what  amount  of  tax  must 
each  town  pay  ?  Ans.    673  e^^jyV'V  dollars. 

29.  In  1850  there  were  in  the  United  States  213  college 
libraries,  containing  942321  volumes ;  what  would  be  the 
average  number  of  volumes  to  each  library  ? 

Ans.    442  4^f^  vols. 

30.  The  number  of  post  offices  in  the  United  States  in 
1853  was  22320,  and  the  entire  revenue  of  the  post  office 
department  was  5937120  dollars ;  what  was  the  average 
revenue  of  each  otfice .''  Ans.    266  dollai'S. 


56  SIMPLE   NUMBERS. 

CONTRACTIONS. 
CASE    I. 

81.  When  the  divisor  is  a  composite  number. 

1.  If  3270  dollars  be  divided  equally  among  30  men,  how 
many  dollars  •will  each  receive  ? 

OPERATION.  Analysis.       If    3270   dollars    be    divided 

5)3270  equally  amoilg  30  men,  each  man  will  receive 

— ~  as  many  dollars  as  30  is  contained  times  in 

6)6o4  3270  dollars.     30  may  be  resolved   into  the 

109  Ans.         factors  5  and  6  ;  and  we  may  suppose  the  30 

men  divided  into  5   groups   of  G  men   each ; 

dividing  the  3270  doUars  by  5,  the  number   of  groups,  we  have 

654,  the  number  of  dollars  to  be  given  to  each  group  ;  and  dividing 

the  604  dollars  by  6,  the  number  of  men  in  each  group,  we  have 

109,  the  number  of  dollars  that  each  man  will  receive.     Hence, 

Rule.  Divide  the  dividend  by  one  %f  the  factors,  and  the 
quotient  thus  obtained  hj  another,  and  so  on  if  there  be  more 
than  tu'o  factors,  until  every  factor  has  hccn  made  a  divisor. 
Tlie  last  quotient  will  be  the  quotient  required. 

EXAMPLES    FOR    PRACTICE. 

2.  Divide  3r.90  by  15  =  3  X  5.  Ans.  246. 

3.  Divide  3528  by  24  =  4  X  6.  Ans.  147. 

4.  Divide  7280  by  35  =  5X7.  Ans.  208. 

5.  Divide  6228  by  36  =  6  X  6.  Ans.  173. 

6.  Divide  33642  by  27  r=  3  X  9.  Ans.  1246. 

7.  Divide  153160  by  56  =  7  X  8.  Ans.  2735. 

8.  Divide  15625  by  125  =  5  x  5  X  5.        Ans.  125. 

82.  To  find  the  true  remainder. 

1.  Divide  1143  by  64,  using  the  foctors  2,  8,  and  4,  and  fiud 
the  true  remainder. 

What  are  contractions  ?     Case  I  is  what  ?     Give  explanation.     Rule. 


DIVISION.  67 

OPERATION.  Analysis.    Divid- 

2)1143  ing    1143    by    2,   ^ve 

^TTZ~  ,  have    a    quotient    of 

8)0/1 1  rem.  ,- 1        ,  .    , 

/  5(1,  and  a  remainder 

4)71 3  X  -^  =    G     "  of  1  undivided,  which, 


17--3X8X  2  =  48 


being   a   part   of  the 
given  dividend,  must 


5.3  true  rem.       also  be  a  part  of  the 

true  remainder.     The 

571    being  a  quotient   arising   from  dividing   by  2,  its    units   are 

2  times  as  great  in  value  as  the  units  of  the  given  dividend,  1143. 
Dividing  the  571  by  8,  we  have  a  quotient  of  71,  and  a  remainder 
of  3  undivided.  As  this  3  is  a  part  of  the  571,  it  must  be  mnltipHed 
by  2  to  change  it  to  the  same  kind  of  units  as  the  1.  This  makes  a 
true  remainder  of  6  arising  from  dividing  by  8.  Dividing  the  71  1)y 
4,  we  have  a  quotient  of  17,  and  a  remainder  of  3  undivided.     Tliis 

3  is  a  part  of  the  71,  the  iniits  of  which  are  8  times  as  great  in  vahie 
as  those  of  the  571,  and  the  units  of  the  571  are  2  times  as  great 
in  value  as  those  of  the  given  dividend,  1143  ;  therefore,  to  change 
this  last  remainder,  3,  to  units  of  the  same  value  as  the  dividend, 
we  multiply  it  by  8  and  2,  and  obtain  a  true  remainder  of  48  arising 
from  dividing  by  4.  Adding  the  three  partial  remainders,  we  obtain 
55,  the  true  remainder.     Hence, 

KuLE.  I.  Multipli/  each  loartial  remainder^  except  the  first, 
hj  all  the  preceding  divisors. 

II.  Add  the  several  products  tvith  the  first  remainder,  and 
the  sum  will  be  the  true  remainder. 

EXAMPLES    FOR   PRACTICE. 

Hem. 

2.  Divide    34712  by  42  =  6X7.  20. 

3.  Divide   40137G  by  G4  =  8  X  8.  32. 

4.  Divide  139074  by  72  =r  3  X  4  X  6.  42. 
f).  Divide  9078126  by  90  =  3  X  5  X  6.  6. 
6.  Divide  18730627  by  120  =  4  X  5  X  6. 


67. 


7.  Divide  73G0479  by  96  =  2  X  6  X  8.         63. 

8.  Divide  24726300  by  70  =  2  X  5 -X  7.         60. 

9.  Divide  5610207  by  84  =  7  X  2  X  G.         15. 

Explain  the  process  of  finding  the  true  remainder  when  dividing  by 
the  factors  of  a  composite  number. 


58  SIMPLE   NUMBERS. 

CASE    11. 

S3.    When  tho  divisor  is  10,  100,  1000,  &c. 

1.  Divide  374  acres  of  land  equally  among  10  men;  how 
many  acres  will  each  have  ? 

OPERATION.  Analysis.     Since  -we  have  shown, 

lICloTI-l  ^^^^  ^°  remove  a  figure  one  place 

toward  the  left  by  annexing  a  cipher 

Quotient.       37 4  Rem.        increases  its  value  tenfold,  or  multi- 

or,  37  J*^)-  acres.  plies  it  by  10,  (6§,)  so,  on  the  con- 

trary, by  cutting  off  or  taking  away 
the  right  hand  figure  of  a  numher,  each  of  the  other  figures  is  removed 
one  place  toward  the  right,  and,  consequently,  the  value  of  each  is 
diminished  tenfold,  or  divided  by  10,  (32.) 

For  similar  reasons,  if  we  cut  off  two  figures,  w-e  divide  by 
100,  if  three,  we  divide  by  1000,  and  so  on.     Hence  the 

Rule.  From  iJte  rifjht  hand  of  the  dividend  cut  off  as 
many  figures  as  there  are  ciphers  in  the  divisor.  Under  the 
figures  so  cut  off,  place  the  divisor,  and  the  ichole  will  form  the 
quotient. 

EXAMPLES    FOR   PRACTICE. 

2.  Divide  47  GO  by  10. 

3.  Divide  3G2078    by  100. 

4.  Divide  130G321    by  1000. 

5.  Divide  9700347    by  10000. 
G.  Divide  20371G0310  by  100000. 

CASE  III. 

84.  When  there  arc  ciphers  on  the  right  hand  of 
the  divisor. 

1.   Divide  437GG1  by  800. 

OPERATION.  Analysis.     In  this  example  Ave 

8l00Vi37Gl01  resolve  800  into  the  factors  8  and 

'      ^— ■ .  100,  and  divide  first  by  100,  by  cut- 

547 61  Rem.  tii^nr  ofi'  two  riglit  hand  figures  of  the 


Caro  II  is  what?      Give  explanation.      Rule.     Case  III  is  what? 
Give  cxplaiiation* 


DIVISION.  69 

dividend,  (§3,)  and  \\c  have  a  quotient  of  437G,  and  a  remainder  of 
61.  AVe  next  divide  by  8,  and  obtain  547  for  a  quotient;  and  the 
entire  quotient  is  ijil^^^i^. 

2.    Divide  347 IG  by  900. 

OPERATION.  Analysis.      Dividing 

9100)34711  G  as  in  the  last  example,  we 

77  ,  have  a  quotient  of  38,  and 

38   Quotient.         0,  2d  rem.  ^^^.^  remainders,  16   and 

5  X  100 -j- 16  =  51 G,  true  rem.  5,       Multiplying    5,    the 

o8|^|,  Ans.  last  remainder,   by    100, 

the  preceding  divisor,  and 
adding  IG,  the  first  remainder,  (§2,)  we  have  516  for  the  true  re- 
mainder. But  this  remainder  consists  of  the  last  remainder,  5,  pre- 
fLxcd  to  the  figures  16,  cut  off  from  the  dividend.     Hence, 

8«5,  When  there  is  a  remainder  after  dividing  by  the  sig- 
nificant figures,  it  must  be  prefixed  to  the  figures  cut  ofT  from 
the  dividend  to  give  the  true  remainder ;  if  there  be  no  other 
remainder,  the  figures  cut  off  from  the  dividend  will  be  the 
true  remainder. 

EXAMPLES    FOn    PKACTICE. 

Quotients.        Rem. 

3.  Divide  347 IG  by  900.  38  516 

4.  Divide  1047G34  by  2400.  436         1234 

5.  Divide  47321046  by  45000.  1051       26046 

6.  Divide  2037903176  by  140000.  63176 

7.  Divide  976031425  by  92000.  3425 

8.  Divide  80013T7G321  by  700000.  376321 

9.  Divide  19070367428  by  4160000.  4584     927428 

10.  Divide  379025644319  by  554000000.  89G44319 

11.  The  circumference  of  the  earth  at  the  equator  is  24898 
miles.  How  muny  hours  would  a  train  of  cars  require  to  travel 
that  distance,  going  at  the  rate  of  50  miles  an  hour  ? 

Ans.    497  48. 

12.  The  sum  of  350000  dollars  is  paid  to  an  army  of  14000 
men  ;  what  does  each  man  receive?  Ans.    25  dollars. 

How  is  the  true  rcraamdcr  found  ? 


60  SIMPLE   XUMBERS. 


EXAMPLES    IX    THE    PRECEDING    RULES. 

1.  George  AYashington  was  born  in  1732,  and  lived  67 
years  ;  in  what  year  did  he  die  ?  Ans.    in  1799. 

2.  How  many  dollars  a  day  must  a  man  ppend,  to  use  an 
income  of  1095  dollars  a  year  ?  Ans.    3  dollars. 

3.  If  I  give  141  dollars  for  a  piece  of  cloth  containing  47 
yards,  for  how  much  must  I  sell  it  in  order  to  gain  one  dollar 
a  yard  ?  Ans.    188  dollars. 

4.  A  speculator  who  owned  500  acres,  17  acres,  98  acres, 
and  121  acres  of  land,  sold  325  acres  ;  how  many  acres  had 
he  left?  Ans.    411  acres. 

5.  A  dealer  sold  a  cargo  of  salt  for  2300  dollars,  and  gained 
625  dollars  ;  what  did  the  cargo  cost  him  ? 

Ans.    1675  dollars. 

6.  If  a  man  earn  60  dollars  a  month,  and  spend  45  dol- 
lars in  the  same  time,  how  long  will  it  take  him  to  save  900 
dollars  from  his  earnings  ? 

7.  If  9  persons  use  a  barrel  of  flour  in  87  days,  how  many 
days  will  a  barrel  last  1  person  at  the  same  rate  ? 

Ans.    783  days. 

8.  The  first  of  three  numbers  is  4,  the  second  is  8  times 
the  first,  and  the  third  is  9  times  the  second ;  what  is  their 
sum?  Ans.    324. 

9.  If  2,  2,  and  7  are  three  factors  of  364,  what  is  the 
other  factor?  ^  Ans.    13. 

10.  A  man  has  3  farms  ;  the  first  contains  78  acres,  the 
second  104  acres,  and  the  third  as  many  acres  as  both  the 
others  ;  how  many  acres  in  the  3  farms  ? 

11.  If  the  expenses  of  a  boy  at  school  are  90  dollars  for 
board,  30  dollars  for  clothes,  12  dollars  for  tuition,  5  dollars 
for  books,  and  7  dollars  for  pocket  money,  what  would  be  the 
expenses  of  27  boys  at  the  same  rate  ?      Ans.  3888  dollars. 

12.  Four  chililren  inherited  2250  dollars  each;  but  one 
dying,  the  remaining  three  inherited  the  whole ;  what  was  the 
share  of  each  ?  Ans.   3000  dollars. 


PROMISCUOrS   EXAMPLES.  61 

13.  Two  men  travel  in  opposite  directions,  one  at  the  rate 
of  35  miles  a  clay,  and  the  other  at  the  rate  of  40  miles  a  day  ; 
how  far  a[)art  are  they  at  the  end  of  6  days? 

14.  Two  men  travel  in  the  same  direction,  one  at  the  rate 
of  35  miles  a  day,  and  the  other  at  the  rate  of  40  miles  a 
day ;  how  far  apart  are  they  at  the  end  of  6  days  ? 

15.  A  man  was  45  years  old,  and  he  had  been  married  19 
years;  how  old  was  he  when  married?         Ans.    26  years. 

IG.  Upon  how  many  acres  of  ground  can  the  entire  popu- 
lation of  the  globe  stand,  supposing  that  25000  persons  can 
stand  upon  one  acre,  and  that  the  population  is  1000000000? 

Ans.    40000  acres. 

17.  Add  384,  1562,  25,  and  946;  subtract  2723  from  the 
sum ;  divide  the  remainder  by  97  ;  and  multiply  the  quotient 
by  142  ;  what  is  the  result?  Ans.    284. 

18.  How  many  steps  of  3  feet  each  would  a  man  take  in 
•walking  a  mile,  or  5280  feet?  Ans.    1760  steps. 

19.  A  man  purchased  a  house  for  2375  dollars,  and  ex- 
pended 340  dollars  in  repairs ;  he  then  sold  it  for  railroad 
stock  worth  867  dollars,  and  235  acres  of  western  land  val- 
ued at  8  dollars  an  acre  ;  how  much  did  he  gain  by  the  trade  ? 

Ans.   32  dollars. 

20.  The  salary  of  a  clergyman  is  800  dollars  a  year,  and 
his  yearly  expenses  are  450  dollars;  if  he  be  worth  1350  dol- 
lars now,  in  how  many  years  will  he  be  worth  4500  dollars  ? 

Ans.    9  years. 

21.  How  many  bushels  of  oats  at  40  cents  a  bushel,  must 
be  given  for  1600  bushels  of  wheat  at  75  cents  a  bushel  ? 

Ans.    3000  bushels. 

22.  Bought  325  loads  of  wheat,  each  load  containing  50 
bushels,  at  2  dollars  a  bushel ;  what  did  the  wheat  cost  ? 

23.  If  you  deposit  225  cents  each  week  in  a  savings  bank, 
and  take  out  75  cents  a  week,  how  many  cents  will  you  have 
left  at  the  end  of  the  year  ?  Ans.    7800  cents. 

24.  The  product  of  two  numbers  is  31383450,  and  one  of 
the  numbers  is  4050  ;  what  is  the  other  number? 


62  SliirLE   KUMBEKS. 

25.  The  Illinois  Central  Railroad  is  700  miles  long,  and 
cost  31647000  dollars;  what  did  it  cost  per  mile? 

Atis.    45210  dollars. 

2C.  Wliat  number  is  that,  which  being  divided  by  7,  the 
quotient  multiplied  by  3,  the  product  divided  by  5,  and  this 
quotient  increased  by  40,  the  sum  will  be  100  ?    Ans.    700. 

27.  How  many  cows  at  27  dollars  apiece,  must  be  given 
for  54  tons  of  bay  at  17  dollars  a  ton  ? 

28.  A  mechanic  receives  56  dollars  for  26  days'  work,  and 
spends  2  dollars  u  day  for  the  whole  time  ;  how  many  dollars 
has  he  left  ?  Ans.   4  dollars. 

29.  If  7  men  can  build  a  house  in  98  days,  how  long  would 
it  take  one  man  to  build  it  ?  Ans.    686  daj's. 

oO.  The  number  of  school  houses  in  the  State  of  New 
Yoik,  in  1855,  was  11,137  ;  suppose  their  cash  value  to  have 
been  5,301,212  dollars,  v/hat  would  be  the  average  value? 

Ans.    476  dollars. 

31.  A  cistern  wtose  capacity  is  840  gallons  has  two  pipes  ; 
through  one  pipe  60  gallons  run  into  it  in  an  hour,  and  through 
the  other  39  gallons  run  out  in  the  same  time ;  in  how  many 
hours  Avill  the  cistern  be  fdled  ?  Ans.    40  hours. 

32.  The  average  beat  of  the  pulse  of  a  man  at  middle  age 
is  about  4500  times  in  an  hour  ;  how  many  times  does  it  beat 
in  24  hours?  Ans.    108000  times. 

33.  How  many  years  from  the  discovery  of  America,  in 
1492,  to  the  year  1900? 

34.  According  to  the  census,  Maine  has  31766  square 
miles;  New  Hampshire,  9280;  Vermont,  10212;  Massachu- 
setts, 7800;  Rhode  Island,  1306;  Connecticut,  4674;  and 
New  York,  47(ii)0;  how  many  more  square  miles  has  all 
New  England  than  New  York? 

35.  What  is  the  remainder  after  dividing  62530000  by 
87900?  v(»s.  33100. 

36.  A  pound  of  cotton  has  been  .-pun  into  a  lliread  8  miles 
in  Icugtli  ;  allowing  235  pounds  for  waste,  how  nnniy  ]>(>nnds 
w  ill  it  take  to  spin  a  thread  to  reach  round  the  earth,  suppos- 
ing-the  distance  to  be  25000  miles  ?        Ans.    3360  pounds. 


I 
i 


ri:uMiscrous  kxamplks.  63 

37.  John  has  854G  dollars,  which  is  342  dollars  less  than 

4  times   as   much   as    Charles    has  ;    how  many   dollars  has 
Charles?  Ans.    2222  dollars, 

38.  The  quotient  of  one  number  divided  by  another  is  37, 
the  divisor  24,0,  and  the  remainder  230  ;  what  is  the  divi- 
dend.?  Ans.    0295, 

oil  "What  number  multiplied  by  72084  will  i>r!)duce 
51'jn048?  Ji>s.    72. 

40.  There  are  two  numbers,  the  greater  of  which  is  73 
times  109,  and  their  diliereiice  is  17  times  28;  wliat  is  the  less 
number?  Ans.    7481, 

41.  The  sum  of  two  numbers  is  3G0,  and  the  less  is  114  ; 
what  is  the  product  of  the  two  numbers  ?  Ans.    28044. 

42.  What  number  added  to  2473248  makes  25G87o4? 

Ans.    Oo.lOG. 

43.  A  farmer  sold  35  bushels  of  wheat  at  2  dollars  a  bush- 
el, and  18  cords  of  wood  at  3  dollars  a  cord;  he  received  9 
yards  of  cloth  at  4  dollars  a  yard,  and  the  balance  in  money  ; 
how  many  dollars  did  he  receive  ?  Ans.    88  dollars. 

44.  A  farmer  receives  684  dollars  a  year  for  produce  from 
his  ftirm,  and  his  expenses  are  375  dollars  a  year ;  how  many 
dollars  will  he  save  in  five  years  ? 

45.  The  salt  manufacturer  at  Syracuse  pays  58  cents  for 
wood  to  boil  one  barrel  of  salt,  10  cents  for  boiling,  5  cents  to 
the  state  for  the  brine,  28  cents  for  the  packing  barixd,  and  3 
cents  for  packing  and  weighing,  and  receives  125  cents  from 
the  purchaser  ;  how  many  cents  does  he  make  on  a  barrel  ? 

Ajts.    21  cents. 

4G.    A   company  of   15    persons  purchase   a   township   of 

western  land  for  28G000  dollars,  of  which  sum  one  man  pays 

GOOO  dollars,  and  the  others  the  remainder,  in  equal  amounts  ; 

how  much  does  each  of  the  others  pay  ?     Ans.    20000  dollars. 

47.  If  25G  be  multiplied  by  25,  the  product  diminished  by 
G25,  and  the  remainder  divided  by  35,  what  will  be  the  quo- 
tient ?  A/(S.    1G5. 

48.  Two  men  start  from  different  places,  distant  189  miles, 
and  travel  toward  each  other ;  one  goes  4  miles,  and  the  other 

5  miles  an  hour ;  in  how  manv  hours  will  thev  meet  ? 


64  SIMPLE   NUMBERS. 

GENERAL  PRINCIPLES   OF   DIVISION. 

§6,  The  quotient  in  Division  depends  upon  the  relative 
values  of  the  dividend  and  divisor.  Hence  any  change  in  the 
value  of  either  dividend  or  divisor  must  produce  a  change  in 
the  value  of  the  quotient.  But  some  changes  may  be  produced 
upon  both  dividend  and  divisor,  at  the  same  time,  that 
uill  not  affect  the  quotient.  The  laws  which  govern  these 
changes  are  called  General  Principles  of  Division,  which  we 
will  now  examine. 

I.    54-^9=:G. 

If  we  multiply  the  dividend  by  3,  we  have 

54  X  3-^9=102^9  =  18, 

and  18  equals  the  quotient,  6,  multiplied  by  3.  Hence, 
Multiplying  the  dividend  by  any  number,  multiplies  the  quotient 
by  the  same  number. 

IL    Using  the  same  example,  54  -^  9  zr:  6. 
If  we  divide  the  dividend  by  3  we  have 

i■3^-^9  =  18-^9  =2, 

and  2  =  the  quotient,  6,  divided  by  3.  Hence,  Dividing  the 
dividend  by  any  number^  divides  the  quotient  by  the  same 
number. 

in.    If  we  multiply  the  divisor  by  3,  we  have 
54  _|_  9  X  3  =  54 -^  27  =  2, 

and  2  =  the  quotient,  G,  divided  by  3.  Hence,  Multiplying 
the  divisor  by  any  number,  divides  the  quotient  by  the  same 
number. 

IV.    If  we  divide  the  divisor  by  3,  we  have 

54-^-^  =  54  4-3  =  18, 


Upon  what  docs  the  value  of  the  quotient  depend  ?     AMiat  is  the 
first   general   principle   of   division  ?      Second  ?      Thii-d  ?      Fourth  i 


GENERAL   PRINCIPLES   OF   DIVISION.  65 

and  18  =  the  quotient,  G,  multiplied  by  3,  Hence,  Dividing 
the  divisor  by  any  number,  midtiplics  the  quotient  by  tlie  same 
number. 

V.  If  we  multiply  both  dividend  and  divisor  by  3,  we  have 

54  X  3  -^  9  X  3  z=  1G2  H-  27  =  6. 

Hence,  MuUiplyincj  both  dividend  and  divisor  by  the  same  num- 
ber, does  not  alter  the  value  of  the  quotient. 

VI.  If  we  divide  botli  dividend  and  divisor  by  3,  we  have 
54-^  a  =18  — 3=:  G. 

Hence,  Dividing  both  dividend  and  divisor  by  the  same  num- 
ber, does  not  alter  the  value  of  the  quotient. 

87.  These  six  examples  illustrate  all  the  different  changes 
we  ever  have  occasion  to  make  upon  the  dividend  and  divisor 
in  practical  arithmetic.  The  principles  upon  which  these 
changes  are  based  may  be  stated  as  follows  : 

Prin.  I.  Multiplying  the  dividend  midtiplies  the  quotient ; 
and  dividing  tlie  dividend  divides  the  quotient.  (8®.  I  and  II.) 
Prin.  II.  Multiplying  the  divisor  divides  the  quotient  ;  and 
\  dividing  the  divisor  multiplies  the  quotient.  (8G.  Ill  and  IV.) 
Pkix.  III.  Multiplying  or  dividing  both  dividend  and 
divisor  by  the  same  number,  does  not  alter  the  quotient.  {^& , 
^   V  and  VI.) 

88.  These  three  principles  may  be  embraced  in  one 

■/ 

GENERAL    LAW, 

A  change  in  the  dividend  produces  a  like  change  in  the 
quotient ;  but  a  change  in  the  divisor  produces  an  opposite 
change  in  tlie  quotient. 

Note.  If  a  number  be  multiplied  and  the  product  divided  by  the 
same  number,  the  quotient  will  be  equal  to  the  number  multiplied. 
Thus,  15  X  4  =  CO,  and  60  -e-  4  =  15. 

Fifth  ?  Sixth  ?  Into  how  many  general  principles  can  these  be  con- 
densed r  AVhat  is  the  tirst  ?  Second  ?  Thii-d  ?  In  what  general  law 
are  these  embraced  ? 


66 


PROPERTIES  OP   NUMBERS. 


EXACT  DIVISORS. 

80c  An  Exact  Divisor  of  a  number  is  one  lliat  gives 
a  whole  number  lor  a  quotient. 

As  it  is  frequently  desirable  to  know  if  a  number  has  an  exaet 
divisor,  we  will  present  a  few  directions  that  will  be  of  assistance, 
particularly  in  finding  exact  divisors  of  large  numbers. 

XoTE.  A  number  whose  unit  figure  is  0,  2,  4,  6,  or  8  is  called  an 
Even  Number.  And  a  nvunber  whose  luiit  tigure  is  1,  3,  5,  7,  or  9,  is 
called  an  Udd  Number. 

2  is  an  exact  divisor  of  all  even  numbers. 

4  is  an  exact  divisor  when  it  will  exactly  divide  the  tens 
and  units  of  a  number.  Thus,  4  is  an  exact  divisor  of  2G8, 
756,  1284. 

0  is  an  exact  divisor  of  every  number  whose  unit  figure  is 
0  or  5.     Thus,  5  is  an  exact  divisor  of  20,  955,  and  2840. 

8  is  an  exact  divisor  when  it  will  exactly  divide  the  hun- 
dreds, tens,  and  units  of  a  number.  Thus,  8  is  an  exact 
divisor  of  1728,  5280,  and  2135G0. 

U  is  an  exact  divisor  when  it  will  exactly  divide  the  sum  of 
(lie  digits  of  a  number.  Thus,  in  2486790,  the  sum  of  the 
digits  2  +  4  +  8  +  6  +  7+94-0  =  36,  and  36-^9  =  4. 

10  is  an  exact  divisor  when  0  occupies  units'  place. 

100  when  00  occupy  the  places  of  units  and  tens. 

1000  when  000  occupy  the  places  of  units,  tens,  and  hun- 
dreds, &c. 

A  composite  number  is  an  exact  divisor  of  any  number, 
when  all  its  factors  are  exact  divisors  of  the  same  number. 
Thus,  2,  2,  and  3  are  exact  divisors  of  12  ;  and  so  also  are  4 
(=2X2)  and  6  (=  2  X  3). 

An  even  number  is  not  an  exact  divisor  of  an  odd  luimber. 

If  an  odd  number  is  an  exact  divisor  of  an  even  number, 


"What  is  an  exact  divisor  ?  "What  is  an  even  number?  An  odd  num- 
ber ?  When  is  2  an  exact  divisor  ?  4?  -5?  9?  10?  1.00?  1000? 
When  is  a  eomiiosite  number  an  exmt  divisor  ?  An  even  number  is 
not  an  exact  divisor  of  what  ?  An  odd  number  is  an  exact  divisor  of 
what  ? 


PACTORTXtt    NUMBERS. 


67 


twice  that  odd  number  is  also  an  exact  divisor  of  tho  even 
number.     Thus,  7  is  au  exact  divisor  of  42 ;  so  also  is  7  X  2, 


or  U. 


PRIME    NUMBERS. 

9©,    A  Prime  Ifumber  is  one  tliat  can  not  be  resolved 
or  s(-:[)arated  into  two  or  more  integral  factors. 

For  reference,  and  to  aid  in  determining  the  prime  factors 
of  composite  numbers,  we  give  the  following  :  — 


1 

59 

139 

2 

Gl 

149 

3 

G7 

151 

5 

71 

157 

7 

73 

103 

11 

79 

167 

13 

83 

173 

17 

89 

179 

19 

97 

181 

23 

101 

19] 

29 

103 

193 

31 

107 

197 

37 

109 

199 

41 

113 

211 

43 

127 

223 

47 

131 

227 

53 

137 

229 

PRIM 

E  NU 

MBER 

3  FRO 

M  1  1 

ro  1000. 

233 

337 

439 

557 

653 

769 

883 

239 

347 

443 

563 

659 

773 

887 

241 

349 

449 

569 

661 

787 

907 

251 

353 

457 

571 

673 

797 

911 

257 

359 

461 

577 

677 

809 

919 

203 

367 

4G3 

587 

683 

811 

929 

269 

373 

4G7 

593 

691 

821 

937 

271 

379 

479 

599 

701 

823 

941 

277 

383 

487 

601 

709 

827 

947 

281 

389 

491 

607 

719 

829 

953 

283 

397 

499 

613 

727 

839 

967 

293 

401 

503 

617 

733 

853 

971 

307 

409. 

509 

619 

739 

857 

977 

311 

419 

521 

631 

743 

859 

983 

313 

421 

523 

641 

751 

863 

991 

317 

431 

541 

643 

757 

877 

997 

331 

433 

547 

647 

761 

881 

FACTOllINO  NUilBERS. 

CAsr;  I. 

91.    To    resolve    any   composlto    numljor   into    its 
prime  factors. 


WTiat  is  a  prim  3  number  ?     Li  f.istoriiig  numbers,  Case  I  is  A\'hat  ? 


2 

2772 

2 

138G 

3 

G93 

3 

231 

7 

77 

11 

11 

1 

68  PROPERTIES   OF   NUMBERS. 

1.  "What  are  the  prime  factors  of  2772  ? 

OPEPvATiON.  Analysis.     We  divide  the  given  number  by 

2,  the  least  prime  factor,  and  the  resuh  by  2 ; 
this  gives  an  odd  number  lor  a  quotient,  divisible 
by  the  prime  factor,  o,  and  the  quotient  resulting 
from  this  division  is  also  divisible  by  3.  The 
mxt  quotiont,  77,  we  divide  by  its  least  prime 
factor,  7,  and  we  obtain  the  quotient  11  ;  this  be- 
ing a  prime  number,  the  division  can  not  be  car- 
ried further.  The  divisors  and  last  quotient,  % 
2,  3,  3,  7,  and  11  are  all  the  prime  factors  of  the 
given  number,  2772.     Hence  tlie 

Rule.  Divide  the  given  number  hi/ ani/  prime  factor  ;  di- 
vide the  quotient  in  the  same  manner,  and  so  continue  the 
division  until  the  quotient  is  a  prime  number.  The  several 
divisors  and  the  last  quotient  will  be  the  prime  factors  required. 

Proof.  The  product  of  all  the  prime  factors  will  be  tha 
given  number. 

EXAMPLES    FOR    PRACTICE. 

2.  What  ai-e  the  prime  factors  of  1 1 40  ?     Ans.  2,  2,  3, 5, 19, 

3.  Wliat  are  the  prime  factors  of  29925  ? 

4.  'NMiat  are  tlie  prime  factors  of  2431  ? 

5.  Find  the  prime  factors  of  12G73. 
G.  Find  the  prime  factors  of  2310. 

7.  Find  the  prime  factors  of  2205. 

8.  What  are  the  prime  factors  of  13981  ? 


CASE    II. 

9°3o    To  resolve  a  number  into  all  the  difibrent  sets 
of  factors  possible. 

1.    In  3o  how  many  sets  of  factors,  and  what  arc  they? 
Giro  explanation,    llulo.     Trooi.     Caoc  II  i^  v/kit  f 


36  =<^ 


CANCELLATION.  69 

OPERATION.  Analysis.     ^VritIng  the  36  at 

2  V  18  ^^^  ^^^^  °^  ^^^^  ^^^^  ^^'  ^^®  arrange 

o  y  1  9  all  the  difl'erent  sets  of  factors  into 

,        (."  -which   it  can   be   resolved   under 
each  other,  as  shown  in  the  opera- 

"  ^  "  tion,  and  we  find  that  36  can  be 

2X2X9  resolved  into  8  sets  of  factors. 
2X3X6 
3X3X4 
2X2X3X3 

EXAMPLES    FOR   PRACTICE. 

2.  How  many  sets  of  factors  in  the  nunaber  24  ?  What 
are  they  ?  A)is.    6  sets. 

3.  In  125  how  many  sets  of  factors  ?     What  are  they  ? 

Ans.    2  sets. 

4.  In  40  how  many  sets  of  factors,  and  wliat  are  they  ? 

Ans.    6  sets. 

0.  In  72  how  many  sets  of  factors,  and  what  are  they  ? 

Ans.    15  sets. 

CANCELLATION. 

93.  Cancellation  is  the  process  of  rejecting  equal  fl^ctors 
from  numbers  sustaining  to  each  other  the  relation  of  dividend 
and  divisor. 

It  has  been  shown  (77)  that  the  dividend  is  equal  to  the 
product  of  the  divisor  multiplied  by  the  quotient.  Hence,  if 
the  dividend  can  be  resolved  into  two  factors,  one  of  which  is 
the  divisor,  the  other  factor  will  be  the  quotient. 

1.  Divide  63  by  7. 

OPERATION.  Analysis.    We  see  in 

Divisor,  "^p  X  9      Dividend.  this  example  that  63  is 

composed  of  the  factors  7 

9      Quotient.               ^nd  9,  and  that  the  factor 
7  is  equal  to  the  divisor. 
Therefore  we  reject  the  factor  7,  and  the  remaining  factor,  9,  is  the 
quotient.  

Give  explanation.  What  is  cancellation  ?  Upon  what  principle  is 
it    based  f       Give  first  explanation. 


70 


PROPERTIES    OF   NUMBERS, 


"Whenever  the  dividend  and  divisor  are  each  composite 
numbers,  the  factors  common  to  botli  may  first  be  rejected 
without  altering  tlie  final  result.      (8T5  Prin.  III.) 

2.    What  is  the  quotient  of  24  times  56  divided  by  7  times 
48? 


24  X  56 
7  X  48 


OPERATION. 

;?  X  0  X  S 


Analysis. 
"We   first   in- 
=.  4    ^ns.  dicate  the  op- 

eration to  be 
performed  by 

■writing  the  numbers  -which  constitute  the  dividend  above  a  line,  and 
those  Avhich  constitute  the  divisor  below  it.  Instead  of  multiplying 
24  by  56,  in  the  dividend,  we  resolve  24  into  the  factors  4  and  G, 
and  5G  into  the  factors  7  and  8 ;  and  48  in  the  divisor  into  the  f\ic- 
tors  6  and  8.  "We  next  cancel  the  factors  6,  7,  and  8,  which  are 
common  to  the  dividend  and  divisor,  and  we  have  left  the  factor  4 
in  the  dividend,  which  is  the  quotient. 

Note.     "NMion  all  the  factors  or  numbers  in  the  dividend  are  can- 
celed, 1  should  be  retained. 

©jj.    If  any  two  numbers,  one  in  the  dividend  and  one  in 
the  divisor,  contain  a  common  factor,  we  may  reject  that  factor. 
3.   In  54  times  77,  how  many  times  63  ? 

Analysis.  In  this  example  we  see  that  9  will 
divide  54  and  63  ;  so  we  reject  9  as  a  fitctor  of  54, 
and  retain  the  factor  6,  and  also  as  a  fi^ctor  of  6.3, 
and  retain  the  factor  7.  Again,  7  will  divide  7  in 
the  divisor,  and  77  in  the  dividend.  Dividing 
both  numbers  by  7,  1  will  be  retained  in  the 
divisor,  and  11  in  the  dividend.  Finally,  the 
l)roduct  of  6  X  1 1  =:  66,  the  quotient. 


OrERATION. 

6        11 

$i  X  711 

00 


4.    Divide  25  X  16  X  12  by  10  X  4  X  6  X  7. 


OPERATION. 

Analysis.       In 

5       4^ 

this,  as  in  the  jire- 

^^  X  l'0  X  ^^      5x4 

-Z<1 

=  2f. 

ceding  example,  we 
reject  all   the  fac- 

10X^X0X7        7 

tors  that  are  com- 

^ 

mon  to  both  divi- 

» 

dend   and   divisor. 

Give  second  explanation. 


CANCELLATION.  7i 

and  \ro  have  remaining  the  factor  7  in  the  divisor,  and  the  factors  5  and 
4  hi  the  dividend.     Completing  the  work,  we  have  ^j^  =:  2&,  Aiis. 

From  the  preceding  examples  and  illustrations  we  derive 
the  following 

Rule.  I.  Write  the  numbers  composing  the  dividend  above 
a  horizontal  line,  and  the  numbers  composing  the  divisor 
below  it. 

II.  Cancel  all  the  factors  common  to  both  dividend  and 
divisor. 

III.  Divide  the  product  of  the  remaining  factors  of  the  div- 
idend by  the  product  of  the  remaining  factors  of  tlie  divisor, 
and  ilie  result  will  be  the  quotient. 


Notes.  1.  Kejccting  a  factor  from  any  niunber  is  dividuig  the  number 
by  that  factor. 

2.  When  a  factor  is  canceled,  the  unit,  1,  is  supposed  to  take  its 
pla(?e. 

I  3.    One  fiictor  in  the  dividend  will  cancel  only  one  equal  factor  in  the 
li  divisor. 

I|       4.    If  all   the  factors   or   numbers  of  the  divisor  are  canceled,  the 
!j  product  of  the  remaining  factors  of  the  dividend  Avill  be  the  quotient. 

II  5.   By  manj'  it  is  thought  more  convenient  to  write  the  factors  of 
\  the  dividend  on  the  right  of  a  vertical  line,  and  the  factors  of  the  divisor 

on  the  left. 


EXAMPLES    FOR    PR'ACTICE. 

1.   What  is  the  quotient  of  16  X  5  X  4  divided  by  20  X  8  ? 

FIRST   OPERATION.  SECOND   OPERATION. 


i0' 

4 


o 


2,  Ans. 
2.   Divide  the  product  of  120  X  44  X  6  X  7  by  72  X  33  X  1 4. 

'j  Hule,  ftrst  step  ?  Second  ?  Third  ?  AMiat  is  the  effect  of  rejecting 
a  factor  ?  What  is  the  quotient  when  iill  the  factors  in  the  divisor  are 
canceled  ? 


72 


mOPERTIES   OF   NUMBERS. 


FIRST   OPERATION. 


/ 


^^X^0X44X(>X^_IOX2 


0       3^ 


¥ 


=  6|,   Am. 


SECOND   OPERATION. 


10 


> 

3 

xn 

U 

0 

11 

20 



2 


6|,  -<4ws. 

« 

3.  Divide  the  product  of  33  X  35  X  28  by  11  X  15  X  14. 

Ans.    1 4. 

4.  AVhat  is  the  quotient  of  21  X  H  X  26  divided  by  14  X 
13?  Ans.   33. 

5.  Divide  the  product  of  the  numbers  48,  72,  28,  and  5,  byl 
the  product  of  the  numbers  84,  15,  7,  and  6,  and  give  the! 
resuh.  Ans.    9|. 

6.  Divide  140  X  39  X  13  X  7  by  30  X  7  X  2G  X  21. 

Ans.   4 J. 

7.  What  is  the  quotient  of6GX9Xl8X5  divided  byj 
22  X  6  X  40?  Ans.    10| 

8.  Divide  the  product  of  200  X  3G  X  30  X  21  by  270  X] 
40  X  15  X  14.  Ans.   2. 

9.  Multiply  240  by  56,  and  divide  the  product  by  GO  mul-l 
tiplied  by  28.  Ans.   8. 

10.  The  product  of  the  numbers  18,  6,  4,  and  42  is  to  be 
divided  by  the  product  of  the  numbers  4,  9,  3,  7,  and  6 ;  ■what] 
is  the  result  ?  Ans.    4. 

11.  IIow  many  tons  of  hay,  at  12  dollars  a  Ion,  must  bol 
given  for  30  cords  of  wood,  at  4  dollars  a  cord  ?     Ans.  10  tons.! 


I 


GREATEST   COMMON   DIVISOR.  78 

12.  How  many  firkins  of  buttei',  each  containing  56  pounds, 
at  13  cents  a  poiniLl,  must  be  given  for  4  barrels  of  sugar,  each 
containing  182  pounds,  at  6  cents  a  pound  ?     Ans.    6  firkins. 

13.  A  tailor  bought  5  pieces  of  cloth,  each  piece  containing 
24  yards,  at  3  dollars  a  yard.  How  many  suits  of  clothes,  at 
18  dollars  a  suit,  must  be  made  from  the  cloth  to  pay  for  it? 

Ans.    20  suits. 

14.  How  many  days'  work,  at  75  cents  a  day,  will  pay  for 
115  bushels  of  corn,  at  50  cents  a  bushel?     Ans.   76§  days. 

GREATEST  COMMON  DIVISOR. 

06.  A  Common  Divisor  of  two  or  more  numbers  is  a 
number  that  will  exactly  divide  each  of  them. 

97.  The  Greatest  Common  Divisor  of  two  or  more  num- 
bers is  the  greatest  number  that  will  exactly  divide  each  of 
them. 

Numbers  pi'ime  to  each  other  are  such  as  have  no  common 
divisor. 

Note.  A  common  divisor  is  sometimes  called  a  Common  Measure  } 
and  the  greatest  common  divisor,  the  Greatest  Common  Measure. 

CASE  I. 

98.  "When  the  numbers  are  readily  factored. 

1.   What  is  the  greatest  common  divisor  of  G  and  10  ? 

Ans.   2. 

OPERATION.  Analysis.     We  readily  find  by  inspection 
G  .  .  1 0  that  2  will   divide   both  tlie   given   numbers ; 
"~          ~  hence  2  is  a  common   divisor;    and  since  the 
'_l quotients  3  and  5  have  no  common  factor,  but 


are  prime  to  each  other,  the  common  divisor, 
2,  must  be  the  greatest  common  divisor. 

2.    What  is  the  greatest  common  divisor  of  42,  63,  and  105  ? 

"\Miat  is  a  common  divisor  ?  The  greatest  common  divisor  ?  A 
common  measure  ?  The  greatest  common  measui-e  ?  "WTiat  is  Case  I  ? 
Give  analysis, 

KP.  4 


3 

42  . 

.  OS  . 

.  105 

7 

14. 

.21  . 

.  35 

2 

o 

.   5 

74  PROPERTIES   OF   NUMBERS. 

OPERATION.  Analysis.     AVe  observe  that   3 

vill  exactly  divide  each  of  the  given 
numbers,  and  that  7  will  exactly 
divide  each  of  the  resulting  quo- 
tients. Hence,  each  of  the  given 
numbers  can  be  exactly  divided  by  3 
•^  ^  '  -*■'  -^'***  times  7;  and  these  numbers  must  be 

component  factors  of  the  greatest 
common  divisor.  Now,  if  there  were  any  other  component  factor  of 
the  greatest  common  divisor,  the  quotients,  2, 3,  5,  would  be  exactly 
divisible  by  it.  But  these  quotients  are  prime  to  each  other.  Hence 
3  and  7  are  all  the  component  factors  of  the  greatest  common  dinsor 
sought. 

3.   "What  is  the  greatest  common  divisor  of  28, 140,  and  280? 

Analysis.     "We  first  divide  by  4 ; 

then  the  quotients   by  7.     The  re- 

suking  quotients,   1,  5,  and   10,  are 

prime  to  each  other.     Hence  4  and 

7  are  all  the  component  factors  of 

the  greatest  common  divisor. 
4  X  7  ziz  2^,  Ans. 

From  these  examples  and  analyses  we  derive  the  following 

Rule.     I.    Write  the  numhers  in  a  line,  ivith  a  vertical  line 
at  the  left,  and  divide  hy  any  factor  common  to  all  the  numhers. 

II.  Divide  the  quotients  in  like  manner,  and  continue  the    \ 
division  till  a  set  of  quotients  is  obtained  that  have  no  common- 
factor. 

III.  Multiply  cdl  the  divisors  together,  and  the  product  will 
he  the  greatest  common  divisor  sought. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  12,  36,  60,  72  ? 

A71S.    12. 

2.  "What  is  the  greatest  common  divisor  of  18,  24,  30,  36, 
42?  A71S.    6. 

Rule,  first  step  f     Second  ?     Third  ? 


^Pif^  '' 

on 
28  .. 

:ration. 
140..  280 

7  . . 

35  . .  70 

1  .. 

5  . .  10 

GREATEST   COMMON   DIVISOR.  75 

3.  What  is  the  greatest  common  divisor  of  72,  120,  240, 
384?  ^ns.    24. 

4.  What  is  the  greatest  common  divisor  of  3G,   126,  72, 
21 G?  Ans.    18. 

5.  AVhat  is  the  greatest  common  divisor  of  42  and  112  ? 

Ans.    14. 

6.  "Wliat  is  the   greatest   common  divisor  of  32,  80,  and 
^  25G?  Ans.    16. 

7.  What  is  the  greatest  common  divisor  of  210,  280,  350, 
630,  and  840?  -Ans.    70. 

8.  What  is  the  greatest  common  divisor  of  300,  525,  225, 
and  375  ?  Ans.    75. 

9.  What  is  the  greatest  common  divisor  of  252,  630,  1134, 

and  138G?  Ans.    126.  ^^. 

10.  What  is  the  greatest  common  divisor  of  96  and  544  ?   ^^^k 

Ans.    32.    ^B^ 

11.  What  is  the  greatest  common  divisor  of  468  and  1184? 

Ans.   4. 

12.  What  is  the  greatest  common  divisor  of  200,  625,  and 
150?  Atis.    25. 

CASE   II. 

99.    When  the  numbers  can  not  be  readily  factored. 

As  the  analysis  of  the  method  under  this  case  depends  upon 
]  three  properties  of  numbers  whicli  have  not  been  introduced, 
we  present  them  in  this  place. 

I.  An  exact  divisor  divides  any  number  of  times  its  dividend. 

II.  A  common  divisor  of  two  numbers  is  an  exact  divisor 
of  their  sum. 

III.  A  common  divisor  of  tvrc  numbers  is  an  exact  divisor 
of  their  difference. 


"SMiat  is  Case  IT  ?      V>Tiat  is  the  first  principle  upon  which  it  is 
founded  ?     Second  ?     Third  ? 


76 


PROPERTIES   OF   NUMBERS. 


1.  "What  is  the  greatest  common  divisor  of  84  and  203  ? 


OPERATION. 

203 
168 

35 

28 


7,  Ans. 


m 


Analysis.  We  draw  two  vertical 
lines,  and  place  the  larger  number  on 
the  right,  and  tlie  smaller  number  on 
the  left,  one  line  lower  down.  ^\'e 
then  divide  203,  the  larger  number,  by 
84,  the  smaller,  and  write  2,  the  quo- 
tient, between  the  verticals,  the  prod- 
uct, 168,  opposite,  under  the  greater 
number,  and  the  remainder,  35,  below. 
We  next  divide  84  by  this  remainder, 
writing  the  quotient,  2,  between  the  verticals,  the  product,  70,  on  the 
left,  and  the  new  remainder,  14,  below  the  70.  We  again  divide  the 
last  divisor,  3.5,  by  14,  and  obtain  2  for  a  quotient,  28  for  a  product, 
and  7  for  a  remainder,  all  of  which  we  write  in  the  same  order  as  in 
the  former  steps.  Finally,  dividing  the  last  divisor,  14,  by  the  last 
remainder,  7,  and  we  have  no  remainder.  7,  the  last  divisor,  is  the 
reatest  common  divisor  of  the  given  numbers. 


84 

2 

70 

2 

14 

2 

14 

2 

0 

In  order  to  show  that  the  last  divisor  in  such  a  process  is 
the  greatest  common  divisor,  we  will  first  trace  the  worlv.  in  the 
reverse  order,  as  indicated  by  the  arrow  line  below. 


OPERATION. 


84 


70 


14 


14 


7  divides  the  14,  as  proved 
by  the  last  division ;  it  will 
also  divide  two  times  14,  or  28, 
(I.)  Xow,  as  7  divides  both 
itself  and  28,  it  Avill  divide  35, 
their  sum,  (11.)  It  will  also 
divide  2  times  35,  or  70,  (I ;) 
and  since  it  is  a  common 
divisor  of  70  and  14,  it  must 
divide  their  sum,  84,  which 
is  one  of  the  given  numbers, 
(IT.)  It  will  also  divide  2 
times  84,  or  108,  (I;)  and 
since  it  is  a  common  divisor  of  168  and  35,  it  must  divide  their 
sum,  203,  the  larger  number,  (II.)  Hence  7  is  a  common  diviso)- 
of  the  given  numbers. 

Again,  tracing  the  work  in  the  direct  order,  as  indicated  below,  we 


203 


168 


35 


28 


Give  analysis. 


GREATEST   COMMON   DIVISOR. 


77 


know  that  the  greatest  common  divisor,  whatever  it  he,  must  divide 

2  times  84,  or  168,  (1.)    Then 
since  it  will  divide  both  168 


^' 


y    203 


84 


70 


14 


•^ 


1G8 

35 

28 
7 


and  203,  it  must  divide  their 
difference,  3j,  (III.)  It  will 
also  divide  2  times  3j,  or  70, 
(I ;)  and  as  it  will  divide  both 
70  and  84,  it  must  divide  their 
difference,  14,  (III.)  It  will 
also  divide  2  times  14  or  28, 
(I ;)  and  as  it  will  divide  both 
28  and  35,  it  must  divide  their 
difference,  7,  (III;)  hence,  it 
cannot  he  greater  than  7. 

Thus  we  have  shown, 

1st.    That  7  is  a  common  divisor  of  the  given  numbers. 
2d.    That  their  greatest  common  divisor,  whatever  it  be, 
cannot  be  greater  than  7.     Hence  it  must  be  7. 

Fi'om  this  example  and  analysis,  we  derive  the  following 

Rule.  I.  Draw  two  verticals,  and  write  the  tivo  numbers, 
one  on  each  side,  the  greater  number  one  line  above  the  less. 

ir.  Divide  the  greater  number  hy  the  less,  iv'riting  the  quo- 
tient between  the  verticals,  the  product  under  the  dividend,  and 
the  remainder  beloio. 

III.  Divide  the  less  number  by  the  remainder,  the  last  divisor 
hy  the  last  remainder,  and  so  on,  till  nothing  remains.  The 
last  divisor  will  be  the  greatest  common  divisor  sought. 

IV.  If  more  than  two  numbers  be  given,  first  find  the  greatest 
co7nmon  divisor  of  tivo  of  them,  and  then  of  this  divisor  and  one 
of  the  remaining  mimhers,  aitd  so  on  to  the  last ;  the  last  common 
divisor  found  will  be  the  greatest  common  divisor  of  all  the 
given  numbers. 

Notes.  1.  "WTien  more  than  two  numbers  are  given,  it  is  better  to 
begin  with  the  least  two. 

2.  If  at  any  point  in  the  operation  a  prime  number  occur  as  a  re- 
mainder, it  must  be  a  common  divisor. or  the  given  numbers  have  no 
common  divisor - 

Rule,  first  step  ?  Second  ?  Thu'd  ?  Fourth  ?  'NMiat  relation  have 
numbers  when  their  difference  is  a  prime  number  ? 


78 


PROPERTIES   or   NUMBERS. 


EXAMPLES    FOR    PRACTICE. 

1.     What  ic  the  greatest  commoi:  divisor  of  251  and  5512? 


OPERATION. 


Ans. 


221 

2 

4 

208 

1 
1 

13 

G 

5512 

442 

1092 
884 

208 
13 

78 

7r> 


0 

2.  Find  the  greatest  common  divisor  of  154  and  210. 

"  Ans.    14. 

3.  What  is  the  greatest  common  divisor  of  31G  and  664? 

Ans.    4. 

4.  What  is  the  greatest  common  divisor  of  679  and  1869? 

Ans.    7. 

5.  AVhat  is  the  greatest  common  divisor  of  917  and  1495? 

Ans.    1. 
G.    What  is  the  greatest  common  divisor  of  1313  and  4108? 

Ans.    13. 

7.  What  is  the  greatest  common  divisor  of  1649  and  5423  ? 

Ans.    17. 

The  following  examples  may  be  solved  by  either  of  the  fore- 
going methods. 

8.  John  has  35  pennies,  and  Charles  50 :  how  shall  they 
arrange  them  in  parcels,  so  that  each  boy  shall  have  the  same 
number  in  each  parcel  ?  Ans.    5  in  each  parcel. 

9.  A  speculator  ha^  3  fields,  the  first  containing  18,  the 
second  24,  and  the  third  40  acres,  which  he  wishes  to  divide 
into  the  largest  possible  lots  having  the  same  nnmber  of  acres 
in  each ;  how  many  acres  in  each  lot  ?  Ans.    2  acres. 


MULTIPLES.  79 

10.  A  farmer  had  231  bushels  of  Avheat,  and  273  bushels 
of  oats,  which  he  wished  to  put  into  the  least  number  of  bins 
containing  the  same  luunber  of  bushels,  without  mixing  the 
two  kinds ;  what  number  of  bushels  must  each  bin  hold  ? 

Ans.    21. 

11.  A  village  street  is  332  rods  loner;  A  owns  124  rods 
front,  B  116  rods,  and  C  92  rods;  they  agree  to  divide  their 
land  into  equal  lots  of  the  largest  size  that  will  allow  each 
one  to  form  an  exact  number  of  lots ;  what  will  be  the  width 
of  the  lots  ?  Ans.    4  rods. 

12.  The  Erie  Railroad  has  3  switches,  or  side  tracks,  of  the 
following  lengths:  3013,  2231,  and  2047  feet;  what  is  the 
length  of  the  longest  rail  that  will  exactly  lay  the  ti'ack  on 
each  switch  ?  Ans.   23  feet. 

13.  A  forwarding  merchant  has  2722  bushels  of  wheat, 
1822  bushels  of  corn,  and  1226  bushels  of  beans,  which  he 
"wishes  to  forward,  in  the  fewest  bags  of  equal  size  that  will 
exactly  hold  either  kind  of  grain ;  how  many  bags  will  it 
take  ?  Ans.   2S<S5. 

14.  A  has  120  dollars,  B  240  dollars,  and  C  384  doFlars  ; 
they  agree  to  purchase  cows,  at  the  higliest  price  per  head  that 
will  allow  each  man  to  invest  all  his  money ;  how  many 
cows  can  each  man  purchase?     Ans.   A  5,  B  10,  and  C  16. 

MULTirLES. 

100.  A  Multiple  is  a  number  exactly  divisible  by  a 
given  number ;  thus,  20  is  a  multiple  of  4. 

103,  A  Common  Multiple  is  a  number  exactly  divisible 
by  two  or  more  given  numbers  ;  thus,  20  is  a  common  multiple, 
of  2.  4,  5,  and  10. 

102.  The  Least  Common  Multiple  is  the  least  number 
exactly  divisible  by  two  or  more  given  numbers;  thus,  24  ie 
the  least  common  multiple  of  3,  4,  6,  and  8. 

\Vhr.t  is  a  multiiile  ?    A  comnion  multiple  ?    The  least  common 
multiple  ? 


80  PROPERTIES   OF   NUMBERS. 

103.  From  the  definition  (100)  it  is  evident  that  the 
product  of  two  or  more  numbei's,  or  any  number  of  times  their 
product,  must  be  a  common  multiple  of  the  numbers.  Hence, 
A  common  multiple  of  two  or  more  numbers  may  be  found  by 
multiplying  the  given  numbers  together, 

104.  To  find  the  least  common  multiple. 

FIRST    METHOD. 

From  the  nature  of  prime  numbers  we  derive  the  follow- 
ing principles :  — 

I.  If  a  number  exactly  contain  another,  it  will  contain  all 
the  prime  factors  of  that  numbei*. 

II.  If  a  number  exactly  contain  two  or  more  numbers,  it 
will  also  contain  all  the  prime  factors  of  those  numbers. 

III.  Tlie  least  number  that  will  exactly  contain  all  the 
prime  factors  of  two  or  more  numbers,  is  the  least  common 
multiple  of  those  numbers. 

1.    Find  the  least  common  multiple  of  30,  42,  C6,  and  78. 

*  ♦ 

OPERATION.  Analysis.      The 

gQ  — -  2  X  3  X  5  number  cannot  be 

JO  _.  2  X  3  X  7  ^^^^  ihan  78,  since 

^P 9VSV11  ^*  must  contain  78; 

~           ov^-iQ  hence  it  must  con- 

78  ==  2  X  o  X  li>  tj^j,^  ^j^e  factors  of 

2X  3  X  13  X  11  X  7  X  5  =  30030, J«s.'  '^''^-^  ^  ^^ 

We  here  have  all  the  prime  factors  of  78,  and  also  all  the  factors  of 
66,  except  the  factor  11.     Annexing  11  to  the  series  of  factors, 

2X3X13X11, 
and  we  have  all  the  prime  factors  of  78  and  66,  and  also  all  the 
factors  cf  42  except  the  factor  7.     Annexing  7  to  the  scries  of  factors, 

2X3X  13X  11  X  7, 
and  we  have  all  the  prime  factors  of  78,  66,  and  42,  and  also  all  the 


How  can  a  common  multiple  of  two  or  more  numbers  be  found  ? 
Pirst  principle  derived  from  prmie  numbers?  becoud  ?  Third? 
Give  analysis. 


LEAST   COMMON   MULTIPLE.  Si 

factors  of  30  except  the  factor  5.     Annexing  5  to  the  series  of  Victors, 

2  X  3  X  13  X  11  X  "^  X  5, 

and  we  have  all  the  prime  factors  of  each  of  the  given  numbers  ;  and 
hence  the  product  of  the  series  of  factors  is  a  common  multiple  of 
the  given  numbers,  (II.)  And  as  no  factor  of  this  series  can  be 
omitted  -without  omitting  a  factor  of  one  of  the  given  numbers,  the 
product  of  the  series  is  the  least  common  multiple  of  the  given 
numbers,  (HI.) 

From  this  example  and  analysis  avc  deduce  the  following 

Rule.    I.   Resolve  the  given  numbers  into  their  prime  factors. 

II.  Take  all  the  prime  factors  of  the  lar-gest  number,  and 
I  such  prime  factors  of  the  other  ninnbers  as  are  not  found  in  the 
largest  number,  and  their  product  will  be  the  least  common 
multiple. 

Note.  When  a  prime  factor  is  repeated  in  any  of  the  given  numbers, 
it  must  be  used  as  many  times,  as  a  factor  of  the  multiple,  as  the 
greatest  number  of  times  it  appears  in  any  of  the  given  numbers. 

EXAMPLES    FOR    PRACTICE. 

2.  Find  the  least  common  multiple  of  7,  35,  and  98. 

Ans.    490. 

3.  Find  the  least  common  multiple  of  24,  42,  and  17. 

Ans.    2856. 

4.  What  is  the  least  common  multiple  of  4,  9,  6,  8  ? 

Ans.    72. 

5.  What  is  the  least  common  multiple  of  8,  15,  77,  385? 

Ans.    9240. 

6.  What  is  the  least  common  multiple  of  10,  45,  75,  90  ? 
I  Ans.   450. 

7.  What  is  the  least  common  multiple  of  12,  15,  18,  35  ? 

A»s.    1260. 


Eule,  first  step  ?     Second  ?     "What  caution  is  given  > 
4* 


82 


PROPERTIES   OP   NUMBERS. 


SECOND    METHOD. 


and  12  ? 


What  is  the  least  common  multiple  of  4,  6,  9, 


Analysis.  We  first  -write 
the  given  numbers  in  a  series, 
■with  a  vertical  line  at  the  left. 
Since  2  is  a  factor  of  some  of 
the  given  numbers,  it  must  be 
a  factor  of  the  least  common 
multiple  sought.  Dividing  as 
many  of  th^  numbers  as  are 
divisible  by  2,  we  write  the 


2 

OPERATION. 

4..  G  ..9.. 

12 

2 

2 

..3. 

.9.. 

6 

3 

3  . 

.9.. 

3 

3 

3 

2  X  2  X  3  X  3  =  36,  Ans. 


quotients  and  the  undivided  number,  9,  in  a  line  underneath.  We 
now  perceive  that  some  of  the  numbers  in  the  second  line  contain 
the  factor  2  ;  hence  the  least  common  multiple  must  contain  another 
2,  and  we  again  divide  by  2,  omitting  to  write  down  any  quotient 
when  it  is  1.  We  next  divide  by  3  for  a  like  reason,  and  still  again 
by  3.  By  this  process  we  have  transferred  all  the  factors  of  each 
of  the  numbers  to  the  left  of  the  vertical ;  and  their  product,  36, 
must  be  the  least  common  multiple  sought,  (104,  III.) 

2.    What  is  the  least  common  multiple  of  10,  12, 15,  and  75  ? 

Analysis.  ^Ve  read- 
ily see  that  2  and  5  are 
among  the  factors  of  the 
given  mnnbers,  and  must 
be  factors  of  the  least 
common  multiple ;  hence 
we  divide  every  number 


2,5 

operation. 
10..  12..  15..  75 

2,3 

6..  3.. 15 

5 

5 

300,  Ans. 


2X5X2X3X5 

that  is  divisible  by  either  of  these  factors  or  by  their  product ;  thus, 
we  divide  10  by  both  2  and  5;  12  by  2  ;  15  by  5;  and  1o  by  5. 
We  next  divide  the  second  line  in  like  manner  by  2  and  3 ;  and 
aftenvards  the  third  line  by  5.  By  this  process  wo  collect  the 
factors  of  the  given  numbers  into  groups ;  and  the  product  of  the 
factors  at  the  left  of  the  vertical  is  the  least  common  multiple  souglit. 

3.    What  is  the  least  common  multiple  of  G,  15,  35,  42, 

and  70  ? 


Give  explanation. 


2,  5 


LEAST   COMMON  MULTIPLE.                             83 

OPERATION.  Analysis.    In  this  opcr- 

1 5  .  .  42  . .  70  ation  we  omit  the  6  and  3(5, 

"■"  because  they  are  exactly  con- 

o  .  .    z  . .  iu  tained  in  some  of  tlie  otlier 


3X7X2X0  =  210,  Ans.        given  numbers;   thus,  6  is 

contained  in  42,  and  'So  in 
70 ;  and  ■whatever  Avill  contain  42  and  70  must  contain  6  and  'do. 
Hence  Me  have  only  to  find  the  least  common  multiple  of  the  re- 
maining numbers,  1  j,  42,  and  70. 

From  these  examples  we  derive  the  following 

Rule.  I.  Write  the  numbers  in  a  line,  omitting  any  of  the 
smaller  numbers  that  are  factors  of  the  larger,  and  draio  a 
vertical  line  at  the  left. 

II.  Divide  bij  any  frime  factor,  or  factors,  tlicd  may  he  con- 
tained in  one  or  more  of  tlie  given  immbers,  and  write  the  quotients 
and  undivided  numbers  in  a  line  underneath,  omitting  the  Vs. 

III.  In  like  manner  divide  the  quotients  and  undivided  num- 
bers, and  continue  the  process  fill  all  tlie  factors  of  the  given 
numbers  have  been  transferred  to  the  left  of  tlie  vertical.  Then 
multiply  these  factors  together,  and  their  product  ivill  be  the  least 
common  multiple  required. 

EXAMPLES    FOR    rRACTICE. 

4.  What  is  the  least  common  multiple  of  12,  15,  42,  and 
60?  Ans.    420. 

5.  What  is  the  least  common  multiple  of  21,  35,  and  42  ? 

Ans.    210. 

6.  "What  is  the  lea^^t  common  multiple  of  25,  60,  100,  and 
125?  Ans.    1500. 

7.  What  is  the  least  common  multiple  of  IG,  40,  9G,  and 
105?  Ans.    3360. 

8.  What  is  the  least  common  multiple  of  4,  16,  20,  48,  GO, 
H    and  72?  Ans.   720. 

9.  What  is  the  least  common  multiple  of  84,  100,  224,  and 
300?  Ans.    16800. 

Rule,  first  step  ?     Second  ?     Third  ? 


84  PROPERTIES    OF   NUMBERS. 

10.  What  is  the  least  common  multiple  of  270,  189,  297, 
243?  A71S.    187110. 

1 1.  What  is  the  least  common  multiple  of  1,  2,  3,  4,  5,  6,  7, 
8,9?  Ans.    2520. 

12.  What  is  the  smallest  sum  of  money  for  Avhich  I  could 
purchase  an  exact  number  of  books,  at  5  dollars,  or  3  dollars, 
or  4  dollars,  or  6  dollars  each  ?  A?7S.    60  dollars. 

13.  A  farmer  has  3  teams;  the  first  can  draw  12  barrels 
of  flour,  the  second  15  barrels,  and  the  third  18  barrels ; 
what  is  the  smallest  number  of  barrels  that  will  make  full 
loads  for  any  of  the  teams?  Ans.    180. 

14.  What  is  the  smallest  sum  of  money  with  Avhich  I  can 
purchase  cows  at  $30  each,  oxen  at  $55  each,  or  horses  at 
$105  each?  A7is.   $2310. 

15.  A  can  shear  41  sheep  in  a  day,  B  63,  and  C  54;  what 
is  the  number  of  sheep  in  the  smallest  flock  that  would  furnish 
exact  days'  labor  for  each  of  them  shearing  alone  ? 

Ans.   15498. 

16.  A  servant  being  ordered  to  lay  out  equal  sums  in  the 
purchase  of  chickens,  ducks,  and  turkeys,  and  to  expend  as 
little  money  as  possible,  agreed  to  forfeit  5  cents  for  every  fowl 
purchased  more  than  was  necessary  to  obey  orders.  In  the 
market  he  found  cliickens  at  12  cents,  ducks  at  30  cents,  and 
turkeys  at  two  prices,  75  cents  and  90  cents,  of  which  he  im- 
prudently took  the  cheaper;  how  much  did  he  thereby  for- 
feit ?  Ans.   80  cents. 


CLASSIFICATION   OF   NUMBERS. 

Numbers  may  be  classified  as  follows : 
BOO.    I.   As  Uven  and  Odd. 
107.    II.    As  Prime  and  Composite. 

What  is  the  first  classification  of  riTimbcrs  ?  What  is  an  even  num- 
ber ?  An  odd  number  ?  JBecond  classification  ?  A  prime  number  ? 
A  composite  number  ? 


CLASSIFICATION   OP   NUMBERS.  85 

108.    III.    As  Integral  ax\(\.  Fractional. 

An  Integral  Number,  or  Integer,  expresses  whole  things. 
Thus,  281 ;  78  boys;  1000  books. 

A  Fractional  Number,  or  Fraction,  expresses  equal  parts 
of  a  thing.  Thus,  half  a  doUar;  three-fourths  of  an  hour; 
seven-eighths  of  a  mile. 

100.      IV.   As  Abstract  and  Concrete. 

ISO.    V.    As  Simple  and  Compound. 

A  Simple  Number  is  either  an  abstract  number,  or  a 
concrete  number  of  but  one  denomination.  Thus,  48,  926; 
48  dollars,  926  miles. 

A  Compound  Number  is  a  concrete  number  whose  value  is 
expressed  in  two  or  more  different  denominations.  Thus,  32 
dollars  15  cents ;  15  days  4  hours  25  minutes ;  7  miles  82 
rods  9  feet  6  inches. 

113.    VI.   As  Like  and  Unlike. 

Like  Numbers  are  numbers  of  the  same  unit  value. 

If  simple  numbers,  they  must  be  all  abstract,  as  6,  62,  487 ; 
or  all  of  one  and  the  same  denomination,  as  5  apples,  62  ap- 
ples, 487  apples ;  and,  if  compound  numbers,  they  must  be 
used  to  express  the  same  kind  of  quantity,  as  time,  distance, 
&c.  Tiius,  4  weeks  3  days  16  hours;  1  week  6  days  9 
hours  ;  5  miles  40  rods  ;  2  miles  100  rods. 

TTnlike  Numbers  are  numbers  of  ditferent  unit  values.  Thus, 
75,  140  dollars,  and  28  miles;  4  hours  30  minutes,  and  5 
bushels  1  peck. 


What  is  the  third  classification  ?  "WTiat  is  an  integral  number  ?  A 
fractional  number  ?  "What  is  the  fourth  classification  ?  An  abstract 
number  ?  A  concrete  number  ?  What  is  the  fifth  classification  ?  A 
simple  number  ?  A  compound  number  ?  Sixth  classification  ?  ^Miat 
are  like  numbers  ?     Unlike  numbers  ? 


86 


FRACTIONS. 


FRACTIONS. 


DEFINITIONS,    NOTATION,    AND    NmiERATION. 

It^,  If  a  unit  be  divided  into  2  equal  parts,  one  of  the 
parts  is  called  07ie  half. 

If  a  unit  be  divided  into  3  equal  parts,  one  of  the  parts  ii 
called  07ie  t/drd,  two  of  the  parts  two  thirds. 

If  a  unit  be  divided  into  4  equal  parts,  one  of  the  parts  is 
called  oite  fourth,  two  of  the  parts  two  fourths,  three  of  the 
parts  three  fourths. 

If  a  unit  be  divided  into  5  equal  parts,  one  of  the  parts  is 
called  one  fifth,  two  of  the  parts  two  fifths,  three  of  the  parts 
tJ tree  fifths,  &c. 

The  parts  are  expressed  by  figures ;  thus, 


One  half   is  written 

J. 

One  fifth       is 

written 

i 

One  third            " 

Two  fifths 

(; 

f 

Two  thirds         " 

o 

S 

One  seventh 

<c 

4- 

One  fourth          " 

i 

Three  eight  lis 

u 

Two  fourths       " 

f 

Five  ninths 

(( 

\ 

Three  fourths    " 

3 

5" 

Eight  tenths 

u 

-h 

Hence  we  see  that  the  parts  into  which  a  unit  is  divided  take 
their  name,  and  their  value,  from  the  numher  of  equal  parts 
into  which  tlie  unit  is  divided.  Thus,  if  we  divide  an  orange 
into  2  equal  parts,  the  parts  are  called /m/rw  ;  if  into  3  equal 
parts,  thirds ;  if  into  4  equal  parts,  fourths.  Sec.  ;  and  each 
third  is  less  in  value  than  each  haf  and  each  fourth  less  than 
each  third ;  and  the  greater  the  numher  of  parts,  the  less 
their  value. 

"When  a  tniit  is  divided  into  any  number  of  equal  parts,  one 
or  more  such  parts  is  a  fractional  part  of  the  whole  number, 
and  is  called  a.  fraction.     Hence 

BBSS.  A  Fraction  is  one  or  more  of  the  equal  parts  of  a 
unit. 


Define  a  fraction. 


DEFINITIONS,    NOTATION,    AND    NUMERATION.  §7 

IS<I.  To  write  a  fraction,  two  integers  are  rtnit'.ired,  one 
to  express  the  number  of  parts  into  which  the  whole  number 
is  divided,  and  the  other  to  express  the  number  of  these  parts 
taken.  Thus,  if  one  dollar  be  divided  into  4  equal  parts, 
the  parts  are  called  fourths,  and  three  of  these  parts  are 
called  three  fourths  of  a  dollar.  This  three  fourths  may  be 
written 

3  the  number  of  parts  taken, 

4  the  number  of  parts  into  which  the  dollar  is  divided, 

Il»5.    The  Denominator  is  the  number  below  the  line. 
It  denominates  or  names  the  parts ;  and 
It  shows  how  many  parts  are  equal  to  a  unit. 

IIG.    The  ITumerator  is  the  number  above  the  line. 
It  numerates  or  numbers  the  parts  ;  and 
It  shows  how   many   parts  are  taken  or  expressed  by 
the  fraction. 

I  if.  The  Terms  of  a  fraction  are  the  numerator  and  de- 
nominator, taken  together. 

1S8.  Fractions  indicate  division,  the  numerator  answering 
to  the  dividend,  and  the  denominator  to  the  divisor.     Hence, 

11^.  The  Value  of  a  fraction  is  the  quotient  of  the  nu- 
merator divided  by  the  denominator. 

22®.  To  analyze  a  fraction  is  to  designate  and  describe 
its  numerator  and  denominator.  Thus,  J  is  analyzed  as  fol- 
lows :  — 

4  is  the  denominator,  and  shows  that  the  unit  is  divided 
,    into  4  equal  parts  ;  it  is  the  divisor. 

3  is  the  numerator,  and  shows  that  3  parts  are  taken  ;  it  is 
the  dividend,  or  integer  divided. 

3  and  4  are  the  terms,  considered  as  dividend  and  divisor. 

The  value  of  the  fraction  is  the  quotient  of  3  -r-  4,  or  f . 


How  many  numbers  are  required  to  write  a  fraction  ?  \\Tiy  ?  Do- 
fine  the  denominator.  The  nimierator.  "SYhat  are  the  terms  of  a  frac- 
tion ?     The  value  ?     ^\^lat  is  the  analysis  of  a  fraction  ? 


88  FRACTIONS. 

EXAMPLES    FOR   PRACTICE. 

Express  the  following  fractions  by  ilgurco :  — - 

1.  Seven  eighths. 

2.  Three  hcenty-jifths. 

3.  Nine  one  hundredtlis, 

4.  Sixteen  thirtieths. 

5.  Thirty-one  one  hundred  eighteenths. 

6.  Seventy-five  7iinety-sixths. 

7.  Two  hundred  fifty-four /oz/r  hundred  forty-thirds. 

8.  Eight  nine  hundred  twenty-Jirsts. 

9.  One  thousand  two  hundred  thirty-two  seventy-five  thou- 
sand six  hundredths. 

10.  Nine  hundred  six  two  hundred  forty-three  thousand 
eighty-seconds. 

Read  and  analyze  the  following  fractions  : 

11  9.        7.        5.12.       15.  9      .45.       125 

^'^'     15'     T2"  '     ^a  '    28'     Tb  '     TT2" '     2':i^  '     ¥2  ff* 

19  90.3  25.        450.  25.  12..        72  6_ 

^^-     TiJiJ  '     TOOTF>     T24iT'     T"boCr  »     S'OOTJ'     U4  7b* 

1Q  17.  1  .  91^.  380  ^h. 

-'■'-'•     T(J¥  '     TrrTTTT  '     'STS'^T  '     4'&ai42g"' 

ISl.    Fractions  are  distinguished  as  Proper  and  Improper. 

A  Proper  Fraction  is  one  wliose  numerator  is  less  than  its 
denominator;  its  value  is  less  than  the  unit,  1.  Thus,  t^,  yV' 
T^U'  §f  ^^^  proper  fractions. 

An  Improper  Fraction  is  one  whose  numerator  equals  or 
exceeds  its  denominator ;  its  value  is  never  less  than  the 
unit,  1.     Thus,  \,  §,  -'4^,  -3g^-,  -f-g,  -'^^5°-  are  improper  fractions. 

ISS.  A  Mixed  Number  is  a  number  expressed  by  an  in- 
teger and  a  fraction;  thus,  4J^,  17^|,  Dy^j  arc  mixed  numbers. 

IS!t.  Since  fractions  indicate  division,  all  changes  in  the 
terms  of  a  fraction  will  afiect  the  value  of  that  fraction  according 
to  the  laws  of  division ;  and  we  have  only  to  modify  the  lan- 
guage of  the  General  Principles  of  Division  (97)  liy  substi- 
tuting the  words  numerator,  denominator,  ^n^  fraction,  or  value 


What  is  a  proper   fraction  ?     An   Improper  fraction  ?     A  mixed 
number  ?     "V^Tiat  flo  fractions  indicate  ? 


REDUCTION.  89 

of  the  fraction,  for  the  words  dividend,  divisor,  and  quotient, 
respectively,  and  we  shall  have  the  following 

GENERAL    PRINCIPLES    OF    FRACTIONS. 

124.  Prin.  I.  3Iidtiplying  the  numerator  midtiplies  the 
fraction,  and  dividing  the  numerator  divides  the  fraction. 

Prin.  II.  Multiplying  the  denominator  divides  the  fraction, 
and  dividing  the  denominator  midtiplies  the  fraction. 

Prin.  III.  Multiplying  or  dividing  both  terms  of  the  frac- 
tion by  the  same  number  does  not  alter  the  value  of  the  fraction. 

These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

\^5.  A  change  in  the  numerator  produces  a  like  change 
in  the  value  of  the  fraction  ;  hut  a  change  in  the  denomina- 
tor produces  an  opposite  change  in  the  value  of  the  fraction. 

REDUCTION. 

case  I. 

„       126.    To  reduce  fractions  to  their  lowest  terms. 

\        A  fraction  is  in  its  lowest  terms  when  its  numerator  and  de- 
nominator are  prime  to  each  other ;  that  is,  when  both  terms 
ii   have  no  common  divisor. 

1.    Reduce  the  fraction  ^|  to  its  lowest  terms. 

first  operation.  Analysis.      Dividin?^   both 

I        4  8  —  24  —  12  —  4      j„,         terms  of  a  fraction  by  the  same 

BO  30  15  5'    ^"A-  ,  , 

1  number  does  not  alter  the  value 

\  of  the  fraction  or  quotient,  (124,  III ;)  hence,  we  divide  both 
terms  of  ||,  by  2,  both  terms  of  the  result,  |A,  by  2,  and  both  terms 
of  this  result  by  3.  As  the  terms  of  |  are  prime  to  each  other,  the 
lowest  terms  of  ||  are  |.  We  have,  in  effect,  canceled  all  the  fac- 
tors common  to  the  numerator  and  denominator. 

First  general  principle  ?  Second  ?  Third  ?  General  law  ?  'NMiat 
is  meant  by  reduction  of  fractions  ?  Case  I  is  what  ?  AVTiat  is 
meant  by  lowest  terms  f     Give  analysis. 


90  FRACTIONS. 

SECOND  OPERATION.  In  this  operation  we  have  divided 

29  \  48  __  4     jlfis^  both   terms  of  the  fraction   by  their 

greatest  common  divisor,   (97,)    and 
thus  performed  the  reduction  at  a  single  division.     Hence  the 

Rule.     Cancel  or  reject  all  factors  common  to  both  numera- 
tor and  denominator.     Or, 

Divide  both  terms  by  their  greatest  common  divisor. 

EXAMPLES    FOR    PnACTICE. 

2.  Reduce    ^^f  to  its  lowest  terms.  Ans.    -i. 

3.  Reduce    §|f  to  its  lowest  terras.  Ans.    |. 

4.  Reduce    HJ-  to  its  lowest  terms.  Ans.  |-^. 

5.  Reduce    f  ||  to  its  lowest  terms. 

6.  Reduce  ^'rVVs  to  its  lowest  terms. 

7.  Reduce   f^'^j  to  its  lowest  terms. 

8.  Reduce   -^^^^  to  its  lowest  terms. 

9.  Reduce  -fg^f °-  to  its  lowest  terras.  Ans.  ^|. 

10.  Reduce   If  |§-  to  its  lowest  terras.  Ans.  ^%. 

11.  Reduce  -i^i^js  to  its  lowest  terms.  Ans.  ^l^. 

12.  Express  in  its  simplest  form  the  quotient  of  441  divided 
by  4G2.  Ans.    f^. 

13.  Express  in  its  simplest  form  the  quotient  of  189  di- 


T^:r- 


vided  by  273.  Ans. 

14.    P^xpress  in  its  simplest  form  the  quotient  of  1344  di- 
vided by  1536.  Ans.  f. 

CASE    II. 

127.    To  reduce  an  improper  fraction  to  a  whole 
or  mixed  number. 

1.    Reduce  ^-5*-  to  «i  whole  or  mixed  number. 

OPERATION.  Analysis.    Since 

■.V*-  =  324— 15  =  2IA  =  21-?,  Ans.        ^^  fifteenths    equal 

1,324  fifteenths  are 
equal  to  as  many  times  1  as  \o  is  contained  times  in  .'521,  which  is 
Slj'j  times.     Or,  since  the  numerator  is  a  dividend  and  the  denom- 

Rulc.     Case  11  is  what  ?     Give  explanation. 


REDUCTION.  91 

inator  a  divisor,  (1  !§,)  ■\ve  reduce  the  fraction  to  an  equivalent 
whole  or  mixed  number,  by  dividing  the  numerator,  324,  by  the 
denominator,  15.     Hence  the 

EuLE.     Divide  the  numerator  hy  the  denominator. 

Notes,  1.  "When  the  denominator  is  an  exact  divisor  of  the  numer- 
ator, the  result  ■will  be  a  whole  number. 

2.  In  all  answers  contaLiimg  fractions,  reduce  the  fractions  to  their 
lowest  terms. 

EXAMPLES    FOR   PRACTICE. 

2.  In  Jy^-  of  a  Aveek,  how  many  weeks  ?  Ans.    ]  f^. 

3.  In  -1-^^  of  a  bushel,  how  many  bushels  ?         Ans.    23|. 

4.  In  A  |i  of  a  dollar,  how  many  dollars? 

5.  In  ^-^^-  of  a  pound,  how  many  pounds  ?  Ans.    54t5-. 

6.  Reduce  -f  J^  to  a  mixed  number. 

7.  Reduce  -^^g^-  to  a  whole  number. 

8.  Change  ^If-  to  a  mixed  number.  Ans.    18§. 

9.  Change  ^ff-  to  a  mixed  number. 

10.  Change -S-V^V"^  to  a  mixed  number.         Ans.    1053||-. 

11.  Change  ^^Igj^^  to  a  whole  number.  Ans.    7032. 

CASE   III. 

128.  To  reduce  a  whole  number  to  a  fraction  hav- 
ing a  given  denominator. 

1.    Reduce  46  yards  to  fourths. 

OPEEATiox.  Analysis.     Since  in  1  yard  there  are  4  fourths, 

4g  in  46  yards  there  are  46  times  4  fourths,  wliich  are 

4  184  fourths  z=  l|^.     In  practice  we  multiply  46, 

the  number  of  yards,  by  4,  the  s;iven  denominator, 

4  >  -^'25'  and  taking  the  product,  184,  for  the  numerator  of  a 
fraction,  and  the  given  denominator,  4,  for  the  de- 
nominator, we  have  ^^.    Hence  we  have  the 

Rule.  Mnltiflij  the.  %ohole  number  by  the  given  denominator ; 
take  the  product  for  a  numerator,  under  which  write  the  given 
denominator. 

Rule.     Case  III  is  what  ?     Give  explanation,     llule. 


li 


92  FRACTIONS. 

XoTE.  A  whole  number  is  reduced  to  a  fractional  form  by  writing 
1  under  it  for  a  denominator  ;  thus,  9  ::=  #. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  25  bushels  to  eighths  of  a  bushel.       Ans.   -§a. 

3.  Reduce  G3  gallons  to  fourths  of  a  gallon.       Ans.   -2^2. 

4.  Reduce  140  pounds  to  sixteenths  of  a  pound. 

5.  In  56  dollars,  how  many  tenths  of  a  dollar?    Ans.   ^^ty. 

6.  Reduce  94  to  a  fraction  whose  denominator  is  9. 

7.  Reduce  180  to  seventy-fifths. 

8.  Change  42  to  the  form  of  a  fraction.  Ans,   ^-f-. 

9.  Change  247  to  the  form  of  a  fraction. 

10.  Change  347  to  a  fraction  whose  denominator  shall 
be  14.  Ans.    iff 8. 

CASE   IV. 

129.  To  reduce  a  mixed  number  to  an  improper 
fraction. 

1,    In  5|  dollars,  how  many  eighths  of  a  dollar  ? 

OPEKATION. 

5  3  Analysis.     Since  in  1  dollar  there  are  8  eighths, 

o  in  o  dollars  there  are    5    times   8  eighths,  or   40 

—  eighths,  and  40  eighths  -(-  3   eighths  zr:  43  eighths, 

■*g*)  Ans.     or  ^-,     From  this  operation  we  derive  the  following 

Rule.  MulUphj  the  whole  nnniher  hy  the  denominator  of 
the  fraction  ;  to  the  product  add  the  nionerator,  and  under  the 
sum  write  the  denominator. 

EXAMPLES    FOR    PRACTICE. 

2.  In  4i  dollars,  how  many  half  dollars?  Ans.    f. 

3.  Ir.  71  f  weeks,  how  many  sevenths  of  a  Aveck  ? 

4.  In  341^  acres,  how  many  fourths  ? 

5.  Change  12/j  years  to  twelfths. 
G.  Change  SG/^-  to  an  improper  fi-action. 

7.  Reduce  2\  ,-Jn  to  an  improper  fraction. 

8.  Reduce  225  i|  to  an  improper  fraction. 


Ans. 

X.^£J.. 

Ans. 

-W- 

Ans. 

-^HF- 

Ans. 

Hi^' 

Case  IV  is  what  ?     Give  explanation.     Rule. 


REDUCTION.  93 

9.    In  96/j%  ^low  many  one  hundred  twentieths  ? 

10.  In  1297g\,  how  many  eighty-fourths  ?      Ans.  -LQ|9^. 

11.  What  improper  fraction  will  express  400§^-  ? 

CASE    V. 

ISO.    To  reduce  a  fraction  to  a  given  denominator. 

As  fractions  may  be  reduced  to  lotver  terms  by  division, 
they  may  also  be  reduced  to  higher  terms  by  multiplication  j 
and  all  higher  terms  must  be  multiples  of  the  lowest  terms. 
(iOS.) 

1.  Reduce  f  to  a  fraction  whose  denominator  is  20. 

OPERATION.  Analysis.    We  first  divide  20,  the 

20  -^  4  zz:  5  required  denominator,  by  4,  the  denomi- 

nator of  the  given  fraction,  to  ascertain 
_  ^  '^  ;^  J.5  ^;j5.        if  it  be  a  multiple  of  this  term,  4.     The 
4  ><  5  division  shows  that  it  is  a  multiple,  and 

that  5  is  the  factor  which  must  be  em- 
ployed to  produce  this  multiple  of  4,  We  therefore  multiply  both 
terms  of  |  by  5,  (124,)  and  obtain  if,  the  desired  result    Hence  the 

Rule.  Divide  the  required  denominator  by  the  denominator 
of  the  given  fraction,  and  midtiply  both  terms  of  the  fraction  by 
the  quotient. 

EXAMPLES    FOR    PKACTICE. 

2.  Reduce  |  to  a  fraction  whose  denominator  is  15. 

Ans.   -jfij. 

3.  Reduce  f  to  a  fraction  whose  denominator  is  35. 

4.  Reduce  |f  to  a  fraction  whose  denominator  is  51. 

Ans.    §f. 

5.  Reduce  f  g  to  a  fraction  whose  denominator  is  150. 

6.  Reduce  ^§g  to  a  fraction  whose  denominator  is  3488. 

Ans.    m%' 

7.  Reduce  ^|^  to  a  fraction  whose  denominator  is  1000. 


Case   Y  is   what  ?     How   are  fractions  reduced  to  higher  terms  ? 
AiVTiat  are  all  higher  terms  ?     Give  analysis.     Rule, 


94  FRACTIONS. 

CASE   YI. 

ISA.    To  reduce  two  or  more  fractions  to  a  com- 
mon denominator. 

A  Common  Denominator  is  a  denominator  common  to  two 
or  more  fractions. 

1.  Reduce  ^  and  f  to  a  common  denominator. 

OPERATION.  Analysis.     AVe  multiply  the  terms  of  the 

3  \/  5  first  fraction  by  the  denominator  of  the  second, 

—  z=z  43^  and  the  terms  of  the  second  fraction  by  the 
4X5  denominator  of  the  fii-st,  (1*24.)  This  must  re- 
^  sy  A  ^'^^'^  ^^'^^  fraction  to  the  same  denominator, 

—  —  8  for  each  new  denominator  Avill  be  the  product 
5X4  of  the  given  denominators.     Hence  the 

Rule.    Multiply  the  terms  of  each  fraction  hy  the  denomina- 
tors of  all  the  other  fractions. 

Note.   ^lixcd  numbers  must  iirst  be  reduced  to  improper  fractions. 
EXAMPLES    FOR    PRACTICE. 

2.  Reduce  §,  4,  «ind  f  to  a  common  denominator. 

//«c        16      12      18 
JinS.     ^^,  245  21. 

3.  Reduce  ^  and  |  to  a  common  denominator. 

A„f.       2  7      2  8 

4.  Reduce  |,  -/j,  and  |  to  a  common  denominator. 

A„o       28  8      210      300 
■^nS.     -560,   J605   360' 

5.  Reduce  f ,  |,  §,  and  ^  to  a,  common  denominator. 

G.    Reduce  j^g,  ^,  and  f  to  a  common  denominator. 

Jiic     2*^     H4     9  6, 
jins.    5-3 J,  -4 5 J,  i^-7- 

7.  Reduce  f,  2^,  |,  and  ^  to  a  common  denominator. 

8.  Reduce  1§,  j%,  and  4  to  a  common  denominator. 

/f  „  o        150      2  4      3  2  0 

Cnse  YI  ic  vlmt  ?     "NVJmt  is  a  rommnn  denominator  ?     Give  ana]3'sis. 


REDUCTION.  95 

CASE  VII. 

IS'3.  To  reduce  fractions  to  the  least  common  de- 
nominator. 

The  Least  Common  Denominator  of  two  or  more  fractions 
is  the  least  denominator  to  wliich  they  can  all  be  reduced,  and 
it  must  be  the  least  common  multiple  of  the  lowest  denom- 
inators. 

1.   Reduce  ^,  |,  and  -f'2  to  the  least  common  denominator. 
OPERATION.  Analysis.    We  first  find 


2,3 

9  •> 

"1  " 


Q       g       22  the  least  common  multiple 

An  of  the  given  denominators, 

_JL! which  is  24.     This  must  be 


2  X  3  X  2  X  2  =  24  the  least  common  denom- 

inator   to   which   the   frac- 
Ans.  tions  can  be  reduced,  (III.) 

We  then  multiply  the  terras 
of  each  fraction  by  such  a 
number  as  will  reduce  the  fraction  to  the  denominator,  24.  E.e- 
ducing  each  fraction  to  this  denominator,  by  Case  V,  we  have  the 
answer. 

Since  the  common  denominator  is  already  determined,  it  is 
only  necessary  to  multiply  the  numerators  by  the  multipliers. 
Hence  the  following 

[       Rule.    I.  Find  the  least  common  multiple  of  the  given  de- 
nominators, for  the  least  common^  denominator. 

II.  Divide  this  common  denominator  hy  each  of  the  given 
denominators,  and  multiph/  each  numerator  hy  the  correspond- 
ing quotient.      The  products  toill  he  the  new  numerators. 

examples  for  practice. 

2.  Reduce  j^-g,  f-^,  %l,  and  /^  to^their  least  common  de- 
nominator. Ans.    -'--A    Mr   i-il   t2.^ 

3.  Reduce  i,  f,  ^^g,  /j  to  their  least  common  denominator. 

J^,  C        )fi«!       192        63  32 

-^'"'     3  3  5'  3  3  5'  3J6'   3aff' 


_"^^^mt  is  Case  VIT?     "What  must  be  the  least  common  denominator  ? 
Give  analysis.     Rule,  first  stop.     Second. 


96  FRACTIONS. 

4.  Reduce  f,  ^i,  f ,  and  6  to  their  least  common  denorai- 

5.  Reduce  5^,  2^,  and  1|  to  their  least  common  denomina- 
tor. Ans.    -\S-VS-V- 

6.  Reduce  j\,  f,  f,  and  \  to  their  least  common  denomi- 

rmtnr  >lw<j       i04      231      352      1^4 

naior.  jms.    ^tg,  zi^i  6T6>  big* 

7.  Reduce  f,  |^,  f ,  2|,  and  j^j  to  their  least  common  de- 
nominator. Ans.    ii§,  T^gV,  yVff'  1-6  8'  iVs- 

8.  Change  %,  {^,  3|,  9,  and  |-  to  equivalent  fractions  hav- 
ing the  least  common  denominator. 

9.  Change  f^,  If,  I,  }|,and  6  to  equivalent  fractions  hav- 
ing the  least  common  denominator. 

10.  Change  2/5,  f  J,  4,  1§,  ^^,  and  f  to  equivalent  frac- 
tions having  the  least  common  denominator. 

11.  Reduce  f,  §,  ^,  and  /^  ^0  a  common  denominator. 

12.  Reduce  |,  |-,  2£,  and  ^  to  a  common  denominator. 

13.  Reduce  ||,  -/j,  §,  and  3^  to   equivalent  fractions   hav- 
ing a  common  denominator.  Ans.    ||,  '^^,  §g,  ^^. 

14.  Change  ^j,  f ,  and  |-  to  equivalent  fractions  having  a 
common  denominator.  Ans.    f^Viy  toViT'  tWit- 

15.  Change  ^\,  7^,  f^,  and  5  to   equivalent  fractions  hav- 
ing a  common  denominator.  •      Ans.    ||,  i^^P-,  |g,  3^^-. 

16.  Change  277,  6^,  /jy,  7,  f,  and  1^  to  equivalent  fractions 
having  a  common  denominator. 

ADDITION. 

133.    1.    What  is  the  sum  of  ^,  f ,  f ,  and  |  ? 

OPERATION.  Analysis.    Since  the 

i  +  i  +  i  +  I-  =  -'#  =  2,  ^«s.  g"'en   fractions^  have  a 

common  denominator,  8, 
their  sum  may  be  found  by  adding  their  numerators,  1,  3,  5,  and 
7,  and  placinfj;  the  sum,  16,  over  the  common  denominator.  We 
thus  obtain  -^  =r:  2,  the  required  sum. 

2.  Add  Z^,  ^(j,  tV,  ^s,  and  j%.  Ans.    2^. 

3.  Add  ^,  1^,  -^.j,  -i^,  W,  and  H-  ^^«-    2/7- 

Give  first  explanation. 


ADDITION.  97 

4.  What  is  the  sum  of  /^,  /g,  /^,  ^f,  ^f,  and  fi? 

5.  What  is  the  sum  of  jV^,  j%%,  -rVa,  t¥(J'  ^nd  jQ?  ? 

6.  What  is  the  sum  of  ^^%  ^-^%,  J-|a  ^ |j.,  and  f  f  f  ? 

i34.    1.   What  is  the  sum  off  and  |  ? 

OPERATION.  Analysis.   In 

a  J-  2  —  2  7    110—.  2 7  +  la  =z^l    An<!  whole     numbers 

we  can  add  like 
numbers  only,  or  those  having  the  same  unit  value ;  so  in  fractions 
v.e  can  add  the  numerators  when  they  have  a  common  denominator, 
but  not  otherwise.  As  -|  and  |^  have  not  a  common  denominator, 
we  first  reduce  them  to  a  common  denominator,  and  then  add  the  nu- 
merators, 27  -[-  10  ^  37,  the  same  as  whole  numbers,  and  place  the 
sum  over  the  common  denominator.     Hence  the  following 

Rule.  I.  When  necessary,  reduce  the  fractions  to  a  com- 
mon or  to  their  least  common  denominator. 

II.  Add  the  numerators,  and  place  the  sum  over  the  common 
denominator. 

Note.  If  the  amoimt  be  an  improper  fraction,  reduce  it  to  a  whole 
or  a  mixed  number. 

EXAMPLES    FOR   PRACTICE. 

2.  Add  f  to  |.  Ans.  §|. 

3.  Add  f  to||^.                    .  Ans.  If^. 

4.  Add  f ,  ^,  f  and  T^.  Ans.  Ifi^. 

5.  Add  if,  §J,  and  ^j.  Ans.  1  A/W 

6.  Add  -1^^,  ^9^,  j\,  and  j\.  Ans.  f . 

7.  Add  f  i,  11^,  11,  ^,  and  f .  Ans.  3jt. 

8.  Add  f ,  i,  f ,  A,  ^,  6,  7,  1^  and  fj,.  Ans.  T^VVa- 

9.  Add  7^,  of,  and  lOf. 

OPERATION.  Analysis.  The  sum  of  the  frac- 

^^2^    |_    ij^  tions  |,  f,  and  |isll^;thesumof 

7  _}_  5  _l_  1 0  rr:  22   "  ^^®  integers,   7,  5,  and  10,  is  22  ; 

and    the   sum   of    both   fractions 

Ans.    23  f  J-  and  integers  is  23  i^.     Hence, 

Give  second  explanation  ?     Rule,  first  step.     Second. 
R.P  6 


! 


98  FRACTIONS. 

To  acid  mixed  numbers,  add  the  fractions  and  integers  sep- 
arately^ and  tlien  add  their  sums. 

Note.  If  the  mixed  numbers  are  small,  the}'  may  be  reduced  to  im- 
proper fractions,  and  then  added  after  the  usual  method. 

10.  What  is  the  sum  of  14|,  Zf^,  1§,  and  ^%  ?     Ans.   21^^. 

11.  "What  is  the  sum  of  |,  lyV,  10|,  and  5  ?       Ans.    18/j. 

12.  Yv'hat  is  the  sum  of  17f,  18  j^,  and  202-^  ? 

13.  What  is  the  sum  of  j^g,  j|,  1  ^,  3,  and  J-f  ? 

■     14.    What  is  the  sum  of  12^4,  327  jV,  and  25i?  Ans.   478/^. 
15.   What  is  the  sum  of  ijg,  -fj,  1  Ja»  2j.  and  -iga  ? 

^«5.  mi. 

IC.   What  is  the  sura  of  3^9^,  2i|,  40f,  and  10  J^j  ? 

17.  Bought  3  pieces  of  cloth  containing  125|,  96f,  and 
48|  yards;  liow  many  yards  in  tlie  3  pieces? 

18.  If  it  take  5^  yards  of  cloth  for  a  coat,  3^  yards  for  a 
pair  of  pantaloons,  and  |  of  a  yard  lor  a  vest,  how  many  yards 
will  it  take  for  all  ?  Ans.    1)  Jg. 

19.  A  farmer  divides  his  farm  into  5  fields;  the  first  con- 
tains 26/2-  acres,  the  second  40  if  acres,  the  third  51  f^  acres, 
the  fourth  59|  acres,  and  the  fifth  G2S  acres  ;  how  many  acres 
in  the  farm?  Ans.    2411^. 

2C,  A  speculator  bought  175^  bushels  of  wheat  for  205^ 
dollars,  325f  bushels  of  barley  for  29 Of  dollars,  270ia  bush- 
els of  corn  for  200 j^  dollars,  and  437 /^  bushels  of  oats  for 
15C)|S^  dollars  ;  how  many  bushels  of  grain  did  he  buy,  and  Iiow 
much  did  he  pay  for  the  whole  ?  A       ^  ^-'^'^3^5  bushels. 

"^'  t  859f  a  dollars. 

SUBTRACTION. 

IS.>.    1.    From  /jj  take  f^. 

OPERATION.  Analysis.    Since  the  given 

^Tj  —  Jj  z=z  '^ij  =  -?,  Ans.  fractions  have  a  common  denom- 

inator, 10,  we  find  the  (lifl'erence 
by  subtracting  3,  the  less  numerator,  from  7,  the  greater,  and  write 


How  arc  mixed  numbers  added  ?      Give  note. 


SUBTRACTION.  99 

the  remainder,  4,  over  the  common  denominator,   10.     "We  thus 
obtain  -^\  :=  |,  the  required  difference. 

2.  From  f  take  |.                            *  Aiis. 

3.  From  {4  take  }i.  Ans. 

4.  From  f  ^  take  ^^y.  Ans. 

5.  From  fl  take  ^|.  -^ks. 

6.  From  -/tj^q  take  yVj.  -4??s. 

7.  From  iff  take  ^Jg.  ^ns. 


1 

4' 

J  3 

TO"' 

1 

J* 
6 
27* 


106.    1.   From  #  take  |-. 


OPEKATION.  Analysis. 

8 5.  —  ^2 30  —  3  2^,3  0  —  _2  —  J       /f„c     As  in  -whole 

numbers,  we 
can  subtract  like  numbers  only,  or  those  having  the  same  unit  value, 
so,  we  can  subtract  fractions  only  when  they  have  a  common  de- 
nominator. As  f  and  |  have  not  a  common  denominator,  we  first 
reduce  them  to  a  common  denominator,  and  then  subtract  the 
less  numerator,  30,  from  the  greater,  32,  and  write  the  difference,  2, 
over  the  common  denominator,  36.  We  thus  obtain  j^g  iz::  Jg,  the 
required  difference.     Hence  the  following 


Rule.  I.  Wl/cn  necesscwy,  reduce  the  fractions  to  a 
commoyi  denominator, 

II.  Subtract  the  numerator  of  the  suhtrahend  from  the 
numerator  of  the  minuend,  and  place  the  difference  over  the 
common  denominator. 

EXAMPLES   FOR   PRACTICE. 


_5 
Tff* 

_9 

¥5' 


2.  From  4  take  f .  Ans. 

3.  From  J-f  take  f.  Ans. 

4.  Subtract  /^  from  |.  Ans.   ■^^^. 

5.  Subtract  -j*-^  from  -f^^.  Ans.   4^. 

6.  Subtract  ^a  from  j^ag.  Ans.   -jW. 

7.  Subtract  f^j^/v  from  fg-.  A71S 


637 

iij^g"* 


8.   What  is  the  difference  between  9^  and  2|-  ? 


Give  explanations     Rule,  first  step.  ■  Second. 


100  FRACTIONS. 


OPERATION.  Analysis.     We   first   reduce    the    frac- 

91  rr  9  4j  tional  parts,  ^  and  -|,  to  a  common  denom- 

23  _-  2-9-  inator,  12.      Since  we  cannot  take  ^\  from 

— ~  ^\,  we  add  1  =  if  to  ^\,  wliich  makes  \^, 

G/j  Ans.  and  j\  from  \^  leaves  ^.     We  now  add  1 

to  tiie  2  in  the  subtrahend,  (50,)    and  say> 

3  from  9  leaves  6.     We  thus  obtain  6^^,  the  difference  required. 

Hence,  to  subtract  mixed  numbers,  we  may  reduce  the 
fractional  parts  to  a  common  denominator,  and  then  subtract 
the  fractional  and  integral  parts  separately.     Or, 

We  may  reduce  the  mixed  numbers  to  improper  fractions, 
and  subtract  the  less  from  the  greater  by  the  usual  method. 

9.    From  81  take  3|. 

10.  From  25 1  take  9yV 

11.  From  4|  take  i|. 

12.  Subtract  1}  from  6. 

13.  Subtract  120^9^  from  4501. 

14.  Subtract  j%\  from  3-j-V 

15.  Find  the  difference  between  49  and  75;^. 

16.  Find  the  difference  between  227f  and  196§. 

17.  From  a  cask  of  wine  containing  314-  gallons,  17|-  gal- 
lons were  drawn  ;  how  many  gallons  remained?  Ans.    13|. 

18.  A  farmer,  having  450-^^  acres  of  land,  sold  3043  acres; 
how  many  acres  had  he  left  ?  Ans.    145^^. 

19.  If  flour  be  bought  for  G^  dollars  per  barrel,  and  sold 
for  72  dollars,  what  will  be  the  gain  per  barrel  ? 

20.  From  the  sura  of  f  and  3J-  take  the  difference  of  4J 
and  5|.  Ans.    3||. 

21.  A  man,  having  25 1  dollars,  paid  6^  dollars  for  coal,  2^ 
dollars  for  dry  goods,  and  f  of  a  dollar  for  a  pound  of  tea; 
how  much  had  he  left?  Ans.   $1G|5. 

22.  What  number  added  to  2|  will  make  7^  ?  Ans.  4^^. 

23.  What  fraction  added  to  j-4-  will  make  ^^  ?    Aiis.  -^s. 

In  how  many  ways  may  mixed  numbers  be  subtracted  ?     What  ore   • 
they  ? 


Ans. 

HI 

Ans. 

16tV 

Ans. 

330i|. 

Ans. 

33¥5- 

MULTIPLICATION.  101 

24.  A  gentleman,  having  2000  dollars  to  divide  among  his 
three  sons,  gave  to  the  first  91 2|  dollars,  to  the  second  545^ 
doHars,  and  to  the  third  the  remainder  ;  how  much  did 
the  tliird  receive  ?  Ans.    $i>-i2^2- 

25.  Bought  a  quantity  of  coal  for  loG^^^  dollars,  and  of 
limiher  for  3502  dollars.  I  sold  the  coal  for  184^  dollars,  and 
the  lumber  for  41G2  dollars.     How  much  was  my  whole  gain? 

Ans.    $114jig. 

ItlULTIPLICATIOX. 
CASE    I. 

137.    To  multiply  a  fraction  by  an  integer. 

1.  If  1  yard  of  cloth  cost  f  of  a  dollar,  how  much  will  5 

yards  cost  ? 

oPERATiOiV.  Analysis.     Since  1  yard  cost 

3  \/  5  — -  JL5  — :  33    J^ns.  ^  fourths  of  a   dollar,  5  yards 

will  cost  5  times  3  fourths  of  a 
dollar,  or  15  fourths,  equal  to  3|  dollars.  A  fraction  is  multiplied 
by  multiplying  its  numerator,  (124.) 

2.  If  1  gallon  of  molasses  cost  -^^  of  a  dollar,  how  much 
•will  5  gallons  cost  ? 

OPERATION.  Analysis.      Since   5,   the 

7    v5  —  n —  13     An<t  multiplier,  is  a  factor  of  20,  the 

denominator,  oi  the  multipli- 
cand, we  perform  the  multiplication  by  dividing  the  denominator, 
20,  by  the  multiplier,  5,  and  we  have  ^,  equal  to  If  dollars.  A 
fraction  is  multiplied  by  dividing  its  denominator,  (124.)     Hence, 

Multiplying  a  fraction  consists  in  multiplying  its  numerator, 
or  dividing  its  denominator. 

Note.  Always  divide  the  denominator  when  it  is  exactly  divisible 
by  the  multiplier. 

EXAMPLES    FOR    PRACTICE. 

3.  Multiply  ^  by  5.  Ans.    \^  =  2f 

4.  Multiply  1-3- by  7.  Ans.    If?. 

Case  I  is  what  ?     Give  explanations.  Deduction. 


0. 

Ih: 

iltiply  ^9^- 

by  12 

G. 

Ml 

iltiply  2-\ 

by  G3 

7. 

Mu 

ihii)ly  oi- 

by  9. 

Ol'ERATION. 

5. 

1 

9 

0 

r, 

4. 
45" 

I 

51  = 
-y-  X  9 : 

—"99.  - 

^/«s. 

9  If 

Aiis. 

5fyr. 

Ans. 

2. 

Ans. 

250. 

Ans. 

1G|. 

l02  FRACTIONS. 

A71S.   7f. 
-4;is.    15. 

Analysis.     In  multiply- 
ing a  mixed  number,  we  first 
multiply  the  fractional  part, 
and   then   the   integer,    and 
.  q  ,  add  the  two  products  ;  or  we 

-*         reduce  the  mixed  number  to 
J  an    improper    fraction,    and 

-  then  multiply  it. 

8.  Multiply  7f  by  12. 

9.  Multiply  J-V  by  8. 

10.  Multiply -jtj  by  51. 

11.  Multiply  15"^-  by  16. 

12,  Multiply  4gi  by  22. 

13.  If  a  man  earn  8f^  dollars  a  week,  bow  many  dollars 
will  he  earn  in  12  weeks  ? 

14,  What  will  9  yards  of  silk  cost  at  -f4-  of  a  dollar  per 
yard  ? 

15,  "What  will  27  bushels  of  barley  cost  at  f  of  a  dollar 
per  bushel?  Ans.    23^  dollars. 

CASE   II. 

S38.    To  multiply  an  integer  by  a  fraction. 

1.    At  75  dollars  an  acre,  how  much  will  ^  of  an  acre  of 
land  cost  r 

FIRST   OPERATION.  ANALYSIS.      3  fifths  of  an 

5  )  75     i.rieo  of  an  aero.  acre  will  cost  three  times  as 

much  as  1  fifth  of  an  acre. 
Dividing  lo  dollars  by  5,  we 
have  1j  dollars,  the  cost  of 
■]l  of  an  acre,  which  we  mul- 
tiply by  ;},  and  ol)tain  4.5 
dollars,  the  cost  of  |  of  an  acre. 

Ex]iliiin  tlip  profpss  of  multiplying  mLxed  niunbcrs.     What  is  Case 
II?     (jive  first  explanation. 


15 

cost  of  -i-  of  an  acre, 

3 

45 

S 

MULTIPLICATION. 


103 


SECOND   OPERATION. 
/  0    price  of  1  acre. 
o . 


5  )  225    cost  of  3  acres. 
^nS.     45       "       "    %    of  an  acre. 


Or,  multiplying  the  price 
of  1  acre  by  3,  we  have  the 
cost  of  3  acres  ;  and  as  ^- 
of  3  acres  is  the  same  as 
I  of  1  acre,  Me  divide  the 
cost  of  3  acres  by  5,  and 
we  have  the  cost  of  ^  of  an 
acre,  the  same  as  in  the  fu'st 
operation.      Hence, 

ILiUiphjing  by  a  fraction  consists  in  multiplying  by  the  nu- 
merator and  dividing  by  the  denominator  of  the  multiplier. 

15 

i^±  Note.    By  using  the  vertical  line  and 

1-,  cancellation,  we  shall  shorten,   aiad  com- 

J^ buie  both  operations  m  one. 


2. 

3. 

4. 

5 

6. 

24 


6f 


45,  Ans. 

EXAMPLES    FOR   PRACTICE. 

Multiply  3  by  |.  Ans.  1  ^. 

Multiply  100  by  yV  Ans.  G4f. 

Multiply  105  by  if  Ans.  85. 

Multiply  19  by  J-f.  Ans.  hll. 
Multiply  24  by  6|. 


OPERATION. 


15  =  f  of  24  ; 
144 


Or. 


^ 


u 

53 


3 


159,  Ans 


159,   Ans. 


7.  Multiply  42  by  9^. 

8.  Multiply  80  by  14jV 

9.  jNIultiply  15G  by  2J. 
10.    At  8  dollars  a  bushel,  what  will  -|  of  a  bushel  of  clover 

seed  cost  ? 


Analysis.  We 
multiply  b)'  the  in- 
teger and  fraction 
6eparatcly,and  add 
the  products  ;  or, 
reduce  the  mixed 
number  to  an  im- 
proper fraction, 
and  then  multiply  by  it. 

Ans.  409^. 
Ans.  1165. 
Ans.      108. 


1 


Give  second  explanation.     Note.     Deduction. 


104 


FRACTIONS. 


11.  If  rv  man  travel  36  miles  a  day,  how  many  miles  will 
he  travel  in  lOf  days  ?  Ans.    384  miles. 

12.  If  a  village  lot  be  worth  4.50  dollars,  what  is  -/j  of  it 
■yyorth  ?  ^««'    262j  dollars. 

13.  At  IG  dollars  a  ton,  what  is  the  cost  of  2^  tons  of  hay  ? 

CASE    III. 

139.    To  multiply  a  fraction  by  a  fraction. 

1.   At  f  of  a  dollar  per  bushel,  how  much  will  f  of  a  bushel 


of  corn  cost  ? 

- 

OPERATION. 

Analysis. 

1st  step, 

1    -1-  4  =r  -^rj,  cost  of  J-  of  a  bushel. 

Since  1  bush- 

2d step, 
Whole  work, 

^2^  X  3  =:  j%     "  "   i  "   «      " 

el  cost  1  of  a 
dollar,  1  of  a 
bushel     will 

Q 


'4 


2 


1 


1,  A71S. 


Or         ^    ^  co^t  f  times  f  of  a  dollar,  or  3  times 

J  of  I  of  a  dollar.  Dividing  §  of  a 
dollar  by  4,  we  have  ^^-^  the  cost  of 
^  of  a  bushel.  A  fraction  is  di- 
vided by  multiplying  its  denomina- 
tor, (124.)  Multiplj-ing  the  cost  of  ^  of  a  bushel  by  3,  we  have  ^^^ 
of  a  dollar,  the  cost  of  |  of  a  bushel.  It  will  readily  be  seen  that  we 
have  multiplied  together  the  two  numerators,  2  and  3,  for  a  new 
numerator,  and  the  two  denominators,  3  and  4,  for  a  new  denom- 
inator, as  shown  in  the  whole  work  of  the  operation.  Hence,  for 
multiplication  of  fractions,  we  have  this  general 

Rule.     I.  Reduce    all    integers    and    mixed  numbers    to 
improper  fractions. 

II.     3/ulti/jIi/  together  the  numerators  for  a  new  numerator,  ^\ 
and  the  denominators  for  a  new  denominator.  ^| 

Note.    Cancel  all  factors  common  to  numerators  and  denominators. 


EXAMPLES   FOR   PRACTICE. 


2.  Multiply  f  by  %. 

3  Multiply  I  by  f . 

4.  IMuUiply  h\  hy  f  f. 

5.  Multiply  4i  by  \. 


Ans.  ^. 

Ans.  -/(J. 

Ans.  j%. 

Ans.  3§. 


"WHiat  is  Case  III  ?     Give  explanation.     Ilule,  first  step  ?     Second  ? 
"Wliat  shall  be  done  M'ith  common  factors  ? 


i 


MULTIPLICATION. 


105 


G. 

7. 
8. 

13  V 


"What  is  the  product  of  ^^^  f,  f,  and  ^  ?  Ans.  Jg. 
Wliat  is  the  product  of  Ig,  f,  2,  and  5^  ?  ^rts.  llji, 
"What  is  the  product  of  f  of  -/j,  f  of  §  of  |,  and  ^  of 


OPERATION.  Or, 

1      $      ^      ^      4      $      '7      , 
—  X  — X  — X  — X  — X— =— ,  ^«5. 

10     6     ;sl     $     7t     $    so 

5 


Note.  Fractions  with  the  ■word  of  between  them 
are  sometimes  called  compotmd  fractions.  The  ■\\'ord 
of  is  simply  an  equivalent  for  the  sign  of  multiplica- 
tion, ancl  signifies  that  the  numbers  between  which 
it  is  placed  are  to  be  multiplied  together. 


5   ^ 

2i 

7 

6 

■$ 

^ 

^ 

$ 

7t 

^ 

i 

$ 

$ 

Ans.    U|. 


9.  Multiply  ^^  of  21  by  |  of  7f 

10.  Multiply  f  of  16  by  ^^  of  26§.  Ans.   85^. 

11.  "What  is  the  product  of  3,  J-  of  f ,  and  |  of  3 1-  ? 

12.  What  is  the  value  of  21  times  f  of  |  of  1  ^  ?       Ans.    2. 

13.  What  is  the  value  of  f  of  4-  of  1 1  times  f  of  8  ? 

14.  What  is  the  product  of  12^  multiplied  by  5i  times  GJ  ? 


Ans. 


4G4^V 


^  of  a  dollar. 


15.  At  I  of  a  dollar  per  yard,  what  will  |  of  a  yard  of 
cloth  cost?  Ans. 

1 6.  If  a  man  own  f  of  a  vessel,  and  sell 
what  part  of  the  whole  vessel  will  he  sell  ? 

17.  AVhen  oats  are  worth  g-  of  a  dollar  per  bushel,  what  is 


of  his  share, 


^  of  a  bushel  worth  ? 


18.  What  will  7f  pounds  of  tea  cost,  at  f  of  a  dollar  per 
pound  ?  Ans.    4^g  dollars. 

19.  What  is  the  product  of  9f  by  4|  ? 

9f 

23         2 

Or,  ^'iX^-=       ■ 


A2 
^3- 


395  product  by  4. 


6f 


Ans.    46 


a      2 

«4f. 


X— =  46. 

7t      ^ 


"What  does  "  of"  signify  when  placed  between  two  fractions  ?     A\'Tiat 
is  a  compound  fi-action  ? 

6* 


106  FRACTIONS. 

To  multiply  mixed  numbers  together  we  may  either  mul- 
tiply by  the  integer  and  fractional  part  separately,  and  then 
add  their  products ;  or,  we  may  reduce  both  numbers  to 
improper  fractions,  and  then  multiply  as  in  the  foregoing  i-ule. 

20.  Muhiply  12aby84.  Ans.    108|. 

21.  "What  cost  G|  cords  of  wood,  at  2^  dollars  a  cord  ? 

22.  What  cost  J  of  2^  tons  of  hay,  at  llj^jj  dollars  a  ton  ? 

Ans.    $21-,3g. 

23.  "What  Avill  8|  cords  of  wood  cost,  at  2f  dollars  per 
cord?  Ans.    22i J  dollars. 

24.  What  must  be  paid  for  f  of  GJ-  tons  of  coal,  at  f  of  7^ 
dollars  per  ton  ? 

25.  A  man  owning  |^  of  a  farm,  sold  ^  of  his  share ;  what 
part  of  the  wliole  farm  had  he  left?  Aiis.    4*. 

26.  Bought  a  horse  for  125|  dollars,  and  sold  hin\for  |-  of 
what  he  cost ;  how  much  was  the  loss  ?  Ans.    ^2d^j. 

27.  A  owned  f  of  123|  acres  of  land,  and  sold  f  of  his 
share;  how  many  acres  did  he  sell  ?  A7is.    49/-^. 

28.  If  a  family  consume  1^  barrels  of  flour  a  month,  how 
many  barrels  will  five  such  families  consume  in  4^^f  months  ? 

DmsION. 

CASE    I. 

140.    To  divide  a  fraction  bj  an  integer. 

1.  If  my  hori^e  eat  jj  of  a  ton  of  hay  in  3  months,  what 
part  of  a  ton  will  last  him  1  month  ? 

OPERATION.  Analysis.    If  he  eat  -^^  of  a  ton  in 

-9-  -1-  3  :=  yV,  Ans.  3  months,  in  1  month  he  will  eat  ^  of 

^"g  of  a  ton,  or  ^^  divided  l)y  3.  Since 
a  fraction  is  divided  by  dividing  its  numerator,  (124,)  we  divide 
the  numerator  of  the  fraction,  ^\,  by  3,  and  we  have  ^^j,-,  the  answer. 

2.  If  3  yards  of  ribbon  cost  |  of  a  dollar,  what  will  1  yard 
cost? 


Case  I  is  what  ?     Give  first  explanation. 


DIVISION.  107 

OPERATION.  Analysis.     Here  we  cannot  exactly 

4  _:_  y  ;—  __5.    jl/is,  divide  the  numerator  by  3;  but,  since  a 

fraction  is  divided  by  multiplying  the 
denominator,  (124,)  we  multiply  the  denominator  of  the  fraction, 
|,  by  3,  and  we  have  ^-^,  the  required  result.     Hence, 

Dividing  a  fraction  consists  in  dividimj  its  numerator,  or 
muhiphjimj  its  denominator. 

Note.     AVe  divide  the  numerator  when  it  is  exactly  divisible  by  the 
divisor;  otherwise  we  multiply  the  denominator. 

EXAMPLES    FOR   PRACTICE. 

8.   Divide    f     by    2.  Ans.     f 


4.  Divide   r^^  by    3.  Ans. 

5.  Divide   ^  by    5.  Ans. 

6.  Divide  yVa  by  25. 

7.  Divide  -{^  by  14.  Ans. 


1 

T 
14 
7  5' 


J  3 


I  7 

4¥T' 


8.  Divide  |^  by  21.  .  Ans. 

9.  If  (')  })oundri  of  sugar  cost  f  of  a  dollar,  how  much  will 
1  pound  cost? 

10.  At  7  dollars  a  barrel,  what  part  of  a  barrel  of  flour  can 
be  bought  for  I  of  a  dollar  ?  Ans.    |. 

1 1.  If  a  }'ard  of  cloth  cost  5  dollars,  what  part  of  a  yard 
can  be  bought  for  f  of  a  dollar  ?  Ans.    ^\. 

12.  If  I)  bushels  of  barley  cost  7^  dollars,  how  much  will  1 
bushel  cost  ? 

orERATION.  NoTF..     AVo  reduce  the  mixed  number 

7X  zz:  ^'S-  to   an   improper  fraction,   and  divide  as 

'JL6   ^(j'_.4^    yl;Z5.  ^'-^"^"^• 

13.  If  12  barrels  of  flour  cost  7G|  dollars,  how  much  will 
1  barrel  cost  ? 

OPERATION.  Analysis.     Here  we  first  divide  as  in 

12  )  764-  simple  numbers,  and  we  have  a  remainder 

of  44.     We  reduce  this  remainder  to  an 

" t'  ^^'^S-  improper  fraction,  SA,  which  we  divide  (as 
in  Ex.  1,)  and  annex  the  residt,  |,  to  the  partial  quotient,  6,  and 
■W2  have  6|,  the  rcqua'cd  result. 


Give  second  explanation.     Deduction. 


108  FRACTIONS. 

14.  How  many  times  will  1G|^  gallons  of  cider  fill  a  vessel 
that  holds  3  gallons  ?  Aiis.    o^^- 

15.  If  9  men  consume  J  of  9|  pounds  of  meat  in  a  day, 
how  much  does  each  man  consume  ?        Ajis.    f  of  a  pound. 

16.  A  man  paid  ^OOf^  for  4  cows;  how  much  was  that 
apiece?  Ans.   $24||. 

CASE    II. 

141.    To  divide  an  integer  by  a  fraction. 

1.   At  J  of  a  dollar  a  yard,  how  many  yards  of  cloth  can  be 
bought  for  12  dollars? 

FIRST  OPERATION.  ANALYSIS.     As  many  yards  as  |  of  a 

22  dollar,  the  price  of  1   yard,  is  contained 

I  times  in  1 2  dollars.     Integers  cannot  be  di- 

vided by  fourths,  because  they  are  not  of 


3  )  48  the  same  denomination.    Reducing  12  dol- 

16    yards  lars  to yb?</7As  by  multiplying,  we  have  48 

fourths ;  and  3  fourths  is  contained  m  43 ' 
fourtlis  16  times,  the  required  number  of  yards. 

SECOND   OPERATION.  ANALYSIS.     Here  we  divide  the  integer 

3)22  by  the  numerator  of  the  fraction,  and  mul- 

tiply   the   quotient   by   the    denominator, 

4:  which  produces  the  same  result  as  in  the 

4  first  operation.    Hence, 

1 6   yards. 

Dividing  ht/  a  fraction  consists  in  multiplyinrj  hy  the  denom 
inator,  and  dividing  hy  the  riumerator  oj  the  divisor. 

EXAMPLES    FOR   PRACTICE. 


2. 

Divide    18  by  f. 

3. 

Divide    63  by  /^ 

4. 

Divide    42  by  f. 

5. 

Divide  120  by  /^ 

6.   Divide  316  by  ^^. 


Ans. 

48. 

Ans. 

117. 

Ans. 

49. 

Ans. 

205^. 

Ans. 

877J. 

Case  II  is  what  ?     Give  first  explanation.     Second.     Deduction. 


DIVISION.  109 

7.  How  many  bushels  of  oats,  worth  f  of  a  dollar  per  bushel, 
will  pay  for  f  of  a  barrel  of  flour,  worth  9  dollars  a  barrel  ? 

Ans.    1 5. 

8.  If  ^  of  an  acre  of  land  sell  for  21  dollars,  what  will  an 
acre  sell  for  at  the  same  rate  ?  Ans.    $49. 

9.  When  potatoes  are  worth  f  of  a  dollar  a  bushel,  and 
corn  I  of  a  dollar  a  bushel,  how  many  bushels  of  potatoes  are 
equal  in  value  to  16  bushels  of  coi-n  ?  Ans.    22 J^. 

10.  If  a  man  can  chop  2f  cords  of  wood  in  a  day,  in  how 
many  days  can  he  chop  22  cords  ? 

OPERATION. 

2J  =  -V- 
22  Analysis.    We  reduce  the  mixed  number 

4  to  an  improper  fraction,  and  then  divide  the 

integer   in   the  same  manner  as  by  a  proper 

1 1  )  88  fraction. 

Ans.    8    days. 

11.  Divide  75  by  13f.  A7is.     5^°-. 

12.  Divide  149  by  24|.  Ans.  G/jl. 

13.  A  farmer  distributed  15  bushels  of  corn  amons:  some 
poor  persons,  giving  them  If  bushels  apiece;  among  how 
many  persons  did  he  divide  it  ? 

14.  Divide  f  of  320  by  f  of  9f  Ayis.    25^. 

15.  Bought  ^  of  7^  cords  of  wood  for  ^  of  $32  ;  how  much 
did  1  cord  cost  ?  A7is.    $5^. 

16.  A  father  divided  183  acres  of  land  equally  among  his 
sons,  giving  them  45£  acres  apiece ;  how  many  sons  had  he  ? 

Ans.    4. 

CASE    III. 

1452.    To  divide  a  fraction  by  a  fraction. 

1.  How  many  pounds  of  tea  can  be  bought  for  |^  of  a  dol- 
lar, at  I  of  a  dollar  a  pound  ? 


How  divide  by  a  mixed  number  ?     Case  HI  is  what  ? 


110  FRACTIONS. 

OPERATION.  Analysis.      As 

First  step,  11  X  3  =  f^       -  many  i)oumls  as  | 

fecoiul  step,  31  JL  2  =z  ?^  =  H.  ofa  dollar  is  coii- 

Y\       o       w       3      2]^  tamed  tunes  in  \1 

Who',,  ^v  ,vi- '~~=^ — X  -^  —  =zli,Ans.of  a   dollar.       1    is 

12       3       1^      2       8  contained  in  u,  1^ 

times,  and  ^  is  con- 
tained in  W  3  times  as  many  times  as  1,  or  3  times  ||,  which  is  ^| 
times,  ivhich  is  the  number  of  pounds  that  could  be  bought  at  \  (;f 
a  dollar  per  pound ;  but  |  is  coiitained  but  \  as  many  times  as  i, 
and  f  ^  divided  by  2  gives  ||,  equal  to  1|  times,  or  the  number  oi' 
l)oiuids   that  can  be  bought  at  |  of  a  dollar  per  pound. 

We  see  in  the  operation  that  we  have  multiplied  the  dividend  by 
the  denominator  of  the  divisor,  and  divided  the  result  by  the  numer- 
ator of  the  divisor,  which  is  in  accordance  with  140  for  dividing  a 
fraction.  Hence,  by  inverting  the  terms  of  the  divisor,  the  two 
fractions  will  stand  in  such  relation  to  each  other  that  we  can  mul- 
tiply together  the  two  up])er  numbers  for  the  numerator  of  the  (pio- 
tient,  and  the  two  lower  numbers  for  the  denominator,  as  shown  in 
the  operation.     For  division  of  fractions,  we  have  this  general 

Rule.  I.  Reduce  integers  and  mixed  numbers  to  irrqjroper 
fractions. 

II.  Invert  tlie  terms  of  the  divisor,  and  iirocced  as  in  midti- 

plication. 

* 
NoTr.s.     1.  The  dividend  and  divisor  may  be  reduced  to  a  common 

demminatov,  and  the  nuinerator  of  the  dividend  be  divided  by  the  nii- 

iiicjutor  of  the  divisor  ;   this  will   give  the  same  result  as  the  rule, 

2,  Apply  cancellation  where  practicable. 

EXAMPLES    FOR   PRACTICE. 


2. 

Divide  I  by   £. 

Ans. 

n 

o 

Divide  ^  by    I. 

Alls. 

31. 

•1. 

Divide  4  by  ^%. 

Ans. 

^•]. 

;"). 

Divide  1  by  /j. 

Ans. 

n 

G. 

Divide  1  by  fj. 

Ans. 

he 

b  1' 

7. 

How  many  times 

• 

IS 

^ 

cont! 

lined 

in 

V 

Ans. 

l^r 

K 

J  low  many  times 

is 

•> 

f 

C'onf; 

lined 

in 

]  2  ? 

Ans. 

31. 

Rule,  first  step.     Second.     "Wliat  other  method  is  mentioned  ? 


DIVISION.  Hi 

9.    How  many  times  is  -/g  contiilned  in  ^^  ?      Ans.    2^. 

10.  How  many  times  is  /g^  contained  in  ^§  ? 

11.  How  many  limes  is  ^  of  |^  contained  in  f  of  2J^? 

12.  Wliat  is  the  quotient  of  -^^y  of  4,  divided  by  f  of  3|  ? 

13.  What  is  the  quotient  of  ^  of  ^  of  oG  divided  by  1  ^^ 
times  •?  ?  ^/is.    o-S. 


1-4.    Wliat  is  the  value  of  ^^  ? 

4? 


■^8 
OPERATION. 

oJ._  J  __7    .   3o_;3^        ^  Tliis  example 

7" —       — Ti    :         ^=^      X  — ^^^  ■&>  Ans.  is  only  another 

^ij        8  "         /^       lOp  iorm     for     ex- 

5  •  r   • 

pressmg  divis- 
ion of  fractions ;  it  is  sometimes  called  a  complex  fraction,  and  the 
process  of  performing  the  division  is  called  reducing  a  com jdcx  frac- 
tion to  a  simple  one. 

We  simjily  reduce  the  upper  number  or  dividend  to  an  improper 
fraction,  and  the  lower  number,  or  divisor,  to  an  improper  fraction, 
and  then  divide  as  before. 

p.? 

15.    Vrhat  is  the  value  of  —  ?  Ans.    fa. 

lU 

IG.    What  is  the  value  of '-?  Ans.    20. 

4 

T 

17.  ■  What  is  the  value  of  ?  Ar.s.    ^V-- 

^ 

2  of  3 

18.  What  is  the  value  of  ^^ -?  Ans.      1. 

JL 

2      of     5 

19.  What  is  the  value  of  ~ ^  ?  Ans.      i. 

fof4i 

20.  If  a  horse  cat  f  of  a  bushel  of  oats  in  a  day,  in  how 
many  days  will  ho  eat  5i  bushels?  Ans.    14. 

21.  If  a  man  spend  If  dollars  per  month  fur  tol)acco,  in 
what  time  will  he  spend  lOf  dollars?  Ans.    Gj  months. 

"SMiat  is  a  complex  fraction  ? 


112  FEACTIONS. 

22.  How  many  times  will  4|  gallons  of  campliene  fill  a 
vessel  tliat  holds  i  of  |  of  1  gallon  ?  Ans.    l(>i. 

23.  If  14  acres  of  meadow  land  produce  322  ton.-  of  liav 
bow  many  tons  will  0  acres  produce  ?  Ans.    11  §. 

24.  If  2  yards  of  silk  cost  $3^,  how  much  less  than  SI 7 
will  9  yards  cost  ?  Ans.    S2|. 

25.  If  f  of  a  yard  of  cloth  cost  -f^y  of  a  dollar,  how  much 
will  1  yard  cost  ? 

26.  A  man,  having  $10,  gave  f  of  his  money  for  clover 
seed  at  S3i-  a  bushel;  how  much  did  he  buy?     Aus.    2  bush. 

27.  How  many  tons  of  hay  can  be  purchased  for  SllOy'^i 
at  $0f  per  ton?  Ans.    12 /q. 

PROJIISCUOUS    EXAMPLES. 

1.  Reduce  ^,  |,  f,  and  ^  to  equivalent  fractions  whose  de- 
nominators shall  be  24.  Ans.    if,  |§,  ^j,  ^V 

2.  Change  4  to  an  equivalent  fraction  having  01  for  its 
denominator  Ans.    |f. 

3.  Find  the  least  common  denominator  of  £,  1§,  ^  of  |,  2, 

i  of  I  of  1  J,. 

4.  Add  4^,  J,  I  of  1^,  3,  and  J-J-. 

5.  Find  the  difference  between  f  of  G/j  and  |  of  4,\. 

6.  The  less  of  two  numbers  is  4750*,  and  their  difference 
is  123f  ;  what  is  the  greater  number?  Ans.    4885j'5. 

7.  What  is  the  difference  between  the  continued  products  of 
3,  I,  i  42,  and  3^,  §,4,  f?  Ans.    Sh}. 

4         2J- 

8.  Reduce  the  fractions — and  -^  to  their  simplest  form. 

JL  11  '■ 

9.  What  number  multiplied  by  f  will  produce  1825|  ? 

Ans.   3043  i. 

10.  A  farmer  had  {■  of  his  sheep  in  one  pasture,  ^  in  an- 
other, and  the  remainder,  which  were  77,  in  a  third  jiasture; 
how  many  sheep  had  he  ?  Ans.    140. 

11.  What  will  7^  cords  of  wood  cost  at  l  of  9i  dollars  per 
cord?  Ans.    $24^ J. 


PROMISCUOUS   EXAMPLES.  113 

12.  At  i  of  a  dollar  per  bushel,  how  many  bushels  of  apples 
can  be  bought  for  5 1  dollars  ? 

13.  Paid  $1837|  for  7350^-  bushels  of  oats ;  how  much  was 
that  per  bushel  ?  Ans.   ^  of  a  dollar. 

11,  If  235 J-  acres  of  land  cost  $4725 1,  how  much  will  G28 
acres  cost.?  Ans.    $12G01. 

15.  A  man,  owning  |  of  an  iron  foundery,  sold  J  of  his  share 
for  $51.0  J;  what  was  the  value  of  the  foundery  ?  Ans.  $1055^. 

16.  14^    less  ^"^^^"   is  f  of  ^  of  what  number? 

I'^^ty  Ans.    27. 

17.  A  merchant  bought  4f  cords  of  wood  at  $3^  per  cord, 
and  paid  for  it  in  cloth  at  f  of  a  dollar  per  yard  ;  how  many 
yards  were  required  to  pay  for  the  wood  ? 

18.  How  many  yards  of  cloth,  f  of  a  yard  wide,  will  line 
201  yards,  1^  yards  wide  ?  Ans.    34^. 

1 9.  If  the  dividend  be  |,  and  the  quotient  tj*^,  what  is  the 
divisor  ? 

20.  If  the  sum  of  two  fractions  be  |,  and  one  of  them  be 
^»(j,  what  is  the  other  ?  Ans.   /^y, 

21.  If  the  smaller  of  two  fractions  be  ||,  and  their  differ- 
ence /tj,  what  is  the  greater  ?  ■        Aiis.    |f. 

22.  If  3 1  pounds  of  sugar  cost  33  cents,  how  much  must  be 
paid  for  65^  pounds  ? 

23.  If  324  bushels  of  barley  can  be  had  for  2591  bushels 
of  corn,  how  much  barley  can  be  had  for  2000  bushels  of 
corn  ?  Ans.    2500  bushels. 

24.  A  certain  sum  of  money  is  to  be  divided  among  5  per- 
sons ;  A  is  to  have  |,  B  |,  C  yV'  ^  ?o»  ^^(^  E  the  remainder, 
which  is  20  dollars ;  what  is  the  whole  sum  to  be  divided  ? 

Ans.    $50. 

25.  What  number,  diminished  by  the  difference  between  | 
and  f  of  itself,  leaves  a  remainder  of  34  ?  Ans.    40, 

26.  If  I  of  a  farm  be  valued  at  $1728,  what  is  the  value  of 
the  whole  ? 


114  FRACTIONS. 


27.  Bought  320  sheep  at  $2|-  per  head ;  afterward  bought 
43.3  at  §1^  per  head  ;  then  sold  f  ofthe  whole  number  at  Sl| 
per  head,  and  the  remamder  at  ^2^  ;  did  1  gahi  or  lose,  and 
how  inueh  ?  Ans.    Lo.^t  S44J-. 

28.  h'  o  be  added  to  both  terms  of  the  fraction  -|,  will  its 
value  be  increased  or  diminithed?        Ans.    Increased  y|j. 

2'j.  If  0  be  added  to  both  terms  of  the  fraction  f ,  Avill  its 
value  be  increased  or  diminished?        Ans.   Diminished  ^\. 

30.  How  many  times  can  a  bottle  holding  |^  of  §  of  a  gal- 
lon, be  filled  from  a  demijohn  containing  f  of  If  gallons  ? 

Ans.    7^. 

31.  Bought  I  of  71  cords  of  wood  for  I  of  $32  ;  how  much 
did  1  cord  cost  ? 

32.  Purchased  728  pounds  of  candles  at  1 6|  cents  a  jiound ; 
had  they  been  purchased  for  3|  cents  less  a  pound,  how  many 
pounds  could  have  been  purchased  lor  the  same  money  ? 

Ans.    953^1- 

33.  AYluit  number,  divided  by  If,  will  give  a  quotient  of 
91?  Ans.    123|. 

34.  The  pi-odiict  of  two  numbers  is  6,  and  one  of  them  is 
184G;  what  is  the  other?  Ans.    gfj- 

35.  A  stone  mason  worked  II5  days,  and  after  paying  his 
board  and  other  expenses  with  f  of  his  earnings,  he  had  S20 
left ;  how  much  did  he  receive  a  day? 

3G.  If  f  of  4  tons  of  coal  cost  $5^,  what  will  f  of  2  tons 
cost  ?  Ans.    $5. 

37.  In  an  orchard  £  of  the  trees  arc  apple  trees,  yV  pt'acli 
trees,  and  the  remainder  are  pear  trees,  which  are  20  more  than 
-^  of  the  wliolr  ;  liow  many  trees  in  the  orchard?     Ans.  800. 

38.  A  man  gave  G|  pounds  of  butter,  at  12  cents  a  pound, 
for  ^  of  a  gallon  of  oil ;  how  much  was  the  oil  worth  a  gal- 
lon ?  Ans.    100  cents. 

39.  A  gentleman,  having  27 H  acres  of  land,  sold  ^  of  it, 
and  gave  g  of  it  to  liis  son;  what  was  the  value  of  the  re- 
mainder, at  $o7^  per  acre  ?  Ans.   ^4u77gV 


PROMISCUOUS   EXAMPLES.  115 

40.  A  horse  and  wagon  cost  $270;  the  hor.sc  cost  1]-  times 
as  much  as  the  wagon ;   what  was  the  cost  of  the  wagon  ? 

41.  AVhat  number  tukeii  from  2^-  times  12|-  will  lea^e 
20,^  >  Ans.    Ux. 

42.  A  merchant  bouglit  a  cargo  of  flour  for  $2173^,  and 
sold'it  for  f- 2_  of  tlie  cost,  thereby  losing  I  of  a  dollar  i)er  bar- 
rel;  how  many  barrels  did  he  purchase  ?  Ans.    126. 

4o.  A  and  B  can  do  a  piece  of  work  in  14  days ;  A  can  do 
I  as  much  as  L ;  in  how  many  days  can  each  do  it  ? 

Ans.    A,  32 1  days  ;  B,  24i  days. 

44.  How  many  yards  of  cloth  f  of  a  yard  wide,are  equal 
to  12  yards  f  of  a  yard  Avide  ?  Ans.    1 1  J. 

45.  A,  B,  and  C  can  do  a  piece  of  work  in  5  days ;  B  and 
C  can  do  it  in  8  days  ;  in  Avhat  time  can  A  do  it  ? 

4G.  A  man  put  his  money  into  4  packages;  in  the  fir.-t  ho 
put  |,  in  the  second  i,  in  the  third  -|,  and  in  the  fourth  the  re- 
mainder, which  was  $24  more  than  -jL  of  the  whole  ;  how  much 
money  had  he?  '  Ans.    $720. 

47.  If  $71  will  buy  3^  cords  of  wood,  how  many  cords  can 
be  bought  for  $10i  ?  Ans.    Uh 

48.  How  many  times  is  ^  of  |  of  27  contained  in  |  of  ^  of 
422? 

41).  A  boy  lost  J-  of  his  kite  string,  and  then  added  30  feet, 
when  it  was  just  \  of  its  original  length;  what  was  the  length 
at  first?  Ans.    100  feet. 

50.  Bought  f  of  a  box  of  candles,  and  having  used  \  of 
them,  sold  the  remainder  for  ^\  of  a  dollar;  how  much  would 
a  box  cost  at  the  same  rate  ?  Ans.    $5|-|. 

51.  A  post  stands  ^  in  the  mud,  J-  in  the  water,  and  21  feet 
above  the  water;  what  is  its  lenn^th  ? 

52.  A  father  left  his  eldest  son  -^  of  his  estate,  hrs  youngest 
son  \  of  the  remainder,  and  his  daughter  the  remainder,  who 
received  $17235  less  than  the  youngest  son;  what  wa-^  the 
value  of  the  estate  ?  Ans.   $21114iJ. 


116  DECIMALS. 


DECIIilAL   rr.ACTIOXS. 

I'^IS.  Decimal  Fractions  are  fractions  ■which  have  for 
their  denominator  10,  100,  1000,  or  1  with  any  number  of 
ciphers  annexed. 

XoTF.s.  1.  The  -word  decimal  is  derived  from  the  Latin  decern, 
■which  signifies  ten. 

2.  Decimal  fractions  are  commonlj'  called  decimals. 

3.  Since  ^V  =^  iVo'  lio  =  Tofo'  ^^•'  ^'^  denominators  of  decimal 
fractions  increase  and  decrease  in  a  tenfold  ratio,  the  same  as  simple 
numbers. 

DECIMAL    NOTATION   AND    NUMERATION. 

144:.  Common  Fractions  are  the  common  divisions  of  a 
unit  into  any  number  of  equal  parts,  as  into  halves,  fiftlis, 
twenty-fourths,  &c. 

Decimal  Fractions  are  the  decimal  divisions  of  a  unit,  thus: 
A  unit  is  divided  into  ten  equal  parts,  called  tenths  ;  each  of 
these  tenths  is  divided  into  ten  other  equal  parts  called  hun- 
dredths ;  each  of  these  hundredths  into  ten  other  equal  parts, 
called  thousandths  ;  and  so  on.  Since  the  denominators  of 
decimal  fractions  increase  and  decrease  by  the  scale  of  10,  the 
same  as  simple  numbers,  in  writing  decimals  the  denomina- 
tors may  be  omitted. 

In  simple  numbers,  the  unit,  1,  is  the  starting  point  of 
notation  and  numeration;  and  so  also  is  it  iiv decimals.  We 
extend  the  scale  of  notation  to  the  left  of  units'  place  in 
wrltiii;^'  integers,  and  to  llie  right  of  units'  place  in  writing 
(Irciiiials.  Thus,  the  first  place  at  the  left  of  units  is  tens, 
and  (lie  first  ))lace  at  the  right  of  units  is  tenths  ;  the  second 
place  at  the  left  is  hundreds,  and  the  second  ))laoe  at  the 
right  is  hundredths  ;  the  third  place  at  the  left  is  thousands, 
and  the  third  place  at  the  right  is  thousandths ;  and  so  on. 


What  are   dociTnal   frnctirms?     TTnw  do   they  difTor  from  common 
fractions  r     Ilow  are  they  written  ? 


NOTATION   AND    NUMERATION.  117 

Tlie  Decimal  Point  is  a  period   (  .  ),  which  must  always  be 
placed  before  or  at  the  left  hand  of  the  decimal.     Thus, 

•j^j    is  expressed  .G 

5  4       «  u  r  I 

Tacr  -^i 

Note.  The  decimal  point  is  also  called  the  Separafrix.  This  is  a 
correct  name  for  it  only  when  it  stands  between  the  integral  and  deci- 
mal i^arts  of  the  same  number. 

.5       is  5  tenths,  which    =  Jg.  of  o  units ; 

.05     is  5  hundredths,  "       =  y'^y  of  5  tenths  ; 

.005  is  0  thousandths,  "       =  ^-L.  of  5  hundredths. 

And  universally,  the  value  of  a  figure  in  any  decimal  place 
is  yij  the  value  of  the  same  figure  in  the  next  left  hand  place. 

The  relation  of  decimals  and  integers  to  each  other  is  clear- 
ly shown  by  the  following 

NUMERATION   TABLE. 


-3  J3 

£   a  -S 

t:   2        «  .2 

t«  .  5  5  §  I  ^  I  a 

^  ,Z  J^  S  ^^  ~  -^    5=~   Z  ^ 

475  3.G2418G95 


— V — 


Integers.  Decimals. 

By  examining  this  table  we  see  that 

Tenths  are  expressed  by  one    figure. 

Hundredths  "  "  "  two     figures. 

Thousandths         "  "  "  three       " 

Ten  thousandths  "  "  "  four         " 

And  any  order  of  decimals  by  one  figure  less  than  the  corre- 
sponding order  of  integers. 

145,    Since   the   denominator  of  tenths   is    10,  of   hun- 


^\Tiat  is  the  decimal  point  ?     AMiat  is  it  sometimes  called  ?     "WTiat  is 
the  value  of  a  figure  in  any  decimal  place  ? 


Its  DECDIALS. 

dredths  100,  of  thousands  1000,  and  so  on,  a  decimal  may  be 
expressed  by  writing  tlie  numerator  only  ;  but  in  tiiis  ca<e 
the  numerator  or  decimal  must  always  contain  as  many 
decimal  places  as  are  equal  to  the  number  of  ciphers  in  the 
denominator;  and  the  denominator  of  a  decimal  will  always 
be  the  unit,  1,  witli  as  many  ciphers  annexed  as  are  equal  to 
the  number  of  ligures  in  the  decimal  or  numerator. 
The  decimal  point  must  never  be  omitted. 

EXAMPLES    FOR    PKACTICE. 

1.  Express  in  figures  thirty-eight  hundredths.  ..^ 

2.  AVrite  seven  tenths. 

3.  Write  three  hundred  twenty-five  thousandths. 

4.  "Write  four  hundredths.  Ans.    .04. 

5.  "Write  sixteen  thousandths. 

G.    "Write  seventy-four  hundred-tliousandths.    Ans.  .00074. 

7.  "Write  seven  hundred  forty-five  millionths. 

8.  "Write  four  tliousand  two  hundred  thirty-two  ten-thou- 
sandths. 

9.  "Write  five  liundred  thousand  millionths. 
10.    Read  tlie  following  decimals  : 

.0,5  .681  .9034  .19248 

.24  .024  .0005  .001385 

.G72  .8471  .100248  .1000087 

Note.  To  read  a  decimal,  we  first  numerate  from  left  to  right,  and 
the  name  of  the  right  hand  figure  is  the  name  of  the  denominator.  "We 
tlien  numerate  from  right  to  left,  as  in  Avhole  niuubers,  to  read  the 
numerator. 

I'lS.    A  mixed  number  is  a  number  consisting  of  integers 

and  decimals;  thus,  71.40G   consists  of  tlie  integral  pait,  71, 

and  tlie  decimal  part,  .40G  ;  it  is  read  the  same  as  "tl-i^^^, 

71  and  406  thousandtlis. 

EXAMPLES    FOR    PRACTICE. 

1.  "Write  eighteen,  and  twentj^-seven  thousandths. 

2.  "Write  four  hundred,  and  nineteen  ten-milliontlis. 


How  many  decimal  places  must  there  be  to  express  any  decimal  ? 


NOTATION    AND    NUMERATION.  119 

3.  Write  fifty-four,  and  fifty-four  millionthg. 

4.  Eighty-one,  and  1  ten-thousandth. 

5.  One  hundred,  and  G7  ten-thousandths. 

6.  Eead  the  following  numbers  : 

18.027      100.0067      400.0000019 
81.0001      54.000054      3.03 
75.075        9.2806       40.40404 


S  IT.  From  the  foregoing  explanations  and  illustrations 
■\ve  derive  the  following  important 

PRINCIPLES    OF    DECIMAL    NOTATION    AND    NUJIERATION. 

1.  The  value  of  any  decimal  figure  depends  upon  its  place 
from  the  decimal  point:  thus  .3  is  ten  times  .03. 

2.  Prefixing  a  cii)her  to  a  decimal  decreases  its  value  the 
same  as  dividing  it  by  ten ;  thus,  .03  is  -^^  the  value  of  .3. 

3.  Annexing  a  cipher  to  a  decimal  does  not  alter  its  value, 
since  it  does  not  change  the  place  of  the  significant  figures  of 
the  decimal ;  thus,  -^jy,  or  .6,  is  the  same  as  -x^^,  or  .60. 

4.  Decimals  increase  from  right  to  left,  and  decrease  from 
left  to  right,  in  a  tenfold  ratio  ;  and  therefore  tliey  may  be 
added,  subtracted,  multiplied,  and  divided  the  same  as  whole 
numbers. 

5.  The  denominator  of  a  decimal,  though  never  expressed, 
is  always  the  unit,  1,  with  as  many  ciphers  annexed  as  there 
are  figures  in  the  decimal. 

6.  To  read  decimals  requires  two  numei'ations  ;  first,  from 
units,  to  find  the  name  of  the  denominator,  and  second,  toivards 
units,  to  find  the  value  of  the  numerator. 

148.    Having  analyzed  all  the  principles  upon  which  the 
Avriting  and  reading  of  decimals   depend,  we  will  now  present 
^  these  principles  in  the  form  of  rules. 

KULE    FOR    DECIMAL    NOTATION. 

I.     Write  the  decimal  the  same  as  a  tvhole  number,  placing 

^VTiat  is  the  first  principle  of  decimal  notation  ?  Second  ?  Third  ? 
Fourth  ?     Fifth  ?     SLxth  ?     Rule  for  notation,  first  step  ? 


li 


120  DECIMALS. 

ciphers  where  necessary  to  give  each  signijicant  figure  its  true 
local  value. 

II.   Place  the  decimal  point  before  the  first  figure. 

RULE    FOR    DECIMAL    NUMERATIOX. 

I.  Numerate  from  the  decimal  point,  to  determine  the  de- 
nominator. 

II.  Numerate  towards  the  decimal  point,  to  determine  the 
numerator. 

III.  Read  the  decimal  as  a  whole  number,  giving  it  the  name 
or  denomination  of  the  right  hand  figure. 

EXAMPLES    FOR   PRACTICE. 

1.  Write  425  millionths. 

2.  Write  six  thousand  ten-thousandths. 

3.  Write  one  thousand  eight  hundred  fifty-nine  hundred- 
thousandths. 

4.  Write  260  thousand  8  billionths. 

5.  Read  the  following  decimals : 

.6321  .748243  .2962999 

.5400027  .6'0000000  .00000006 

6.  Write  five  hundred  two,  and  one  thousand  six  millionths. 

7.  Write  thirty-one,  and  two  ten-millionths. 

8.  Write  eleven  thousand,  and  eleven  hundred-thousandths. 

9.  Write  nine  million,  and  nine  billionths. 

10.  Write  one  hundred  two  tenths.  Ans.    10.2. 

11.  Write  one  hundred  twenty-four  thousand  three    hun- 
dred fifteen  thousandths. 

12.  Write  seven  hundred  thousandths. 

13.  Write  seven  hundred-thousandths. 

14.  Read  the  following  numbers: 

12.36  9.052  62.9999 

142.847  32.004  1858.4583 

1.02  4.0005  27.00045 

Becond  ?     Rule  for  numeration,  first  step  ?     Second  ?     Third  f 


EEDUCTION. 


REDUCTIOX. 


121 


CASE    I. 

149.  To  reduce  decimals  to  a  common  denomina- 
tor. 

1.  Reduce  .5,  .375,  3.25401,  and  4G.13  to  their  least  com- 
mon decimal  denominator. 

OPERATION.  Analysis.      The    given    decimals    must    contain 

.50000  3S  many  places   each,  as  are  equal  to  tho  greatest 

87500  number    of    decimal    figures    in    any  of  the    given 

o  9- )Ai  decimals.     Wo    find    tliat    tho    tliird    number  con- 

tains  five  decimal  places,   and   henco  100000   must 

4:U.ioUUU  |jg  ^  common    denominator.     As    annexing   ciphers 

to  decimals  does  not  alter  their  value,  (144.,  3)  we  give  to  each  number 

five  decimal  places  by  annexing  ciphers,   and  thus  reduce  the  given 

decimals  to  a  common  denominator.     Hence, 

Rule.  Give  to  each  number  the  same  number  of  decimal 
places,  by  annexing  ciphers. 

Notes.  1.  If  the  numbers  be  reduced  to  the  denominator  of  that 
one  of  the  given  numbers  having  the  greatest  nimiber  of  decimal  places, 
they  will  have  their  least  common  decimal  denominator. 

2.  A  whole  number  may  readily  be  reduced  to  decimals  by  placing 
the  decimal  point  after  units,  and  annexing  ciphers ;  one  cipher  re- 
ducing it  to  tenths,  two  ciphers  to  hundredths,  three  ciphers  to  thou- 
sandths, and  so  on. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  .17,  24.6,  .0003,  84,  and  721.8000271  to  their 
least  common  denominator. 

3.  Reduce  7  tenths,  24  thousandths,  187  millionths,  5  hun- 
dred millionths,  and  10845  hundredths  to  their  least  common 
denominator. 

4.  Reduce  to  their  least  common  denominator  the  following 
decimals  :  1000.001,  841.78,  2.6004,  90.000009,  and  6000. 

What  is  meant  by  the  reduction  of  decimals?    Case  I  is  what? 
Give  explanation.     Rule. 
R.P  6 


122  DECIMALS. 

CASE    II. 

I*i0.    To  reduce  a  decimal  to  a  common  fraction. 

1.  Reduce  .75  to  its  equivalent  common  fraction. 

OPERATION.  Analysis.     We  omit  the  decimal  point, 

rr  -  7  5    3  supply  the  proper  denominator  to  the  deci- 

^  '^  ^  mal,  and  then  reduce  the  common  fraction 

thus  formed  to  its  lowest  terms.     Hence, 

Rule.      Omit   the  decimal  point,  and  sttpjili/  the  proper 
denominator. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  .12.5  to  a  common  fraction.  Ans.  ^. 

3.  Reduce  .16  to  a  common  fraction.  Ans.  ^. 

4.  Reduce  .655  to  a  common  fraction.  Ans.  ^J^. 

5.  Reduce  .9375  to  a  common  fraction.  Ans.  -J|. 

6.  Reduce  .0008  to  a  common  fraction.  Ans.  xjVu* 

CASE  hi. 
15I.    To  reduce  a  common  fraction  to  a  decimal. 

1.  Reduce  f  to  its  equivalent  decimal. 

FIRST  OPERATION.  ANALYSIS.     We  first  annex 

j3  —  noo —   75   —  7K    Ay,,  the   same   number  of  ciphers 

to  both  terms  oi  the  traction ; 
SECOND  OPERATION.  '  ^^lis  ^locs  not  alter  its  value. 

We  then  divide  both  resulting 
terms  by  I,  the  sifjnificant  fig- 
•75  ure  of  the  denominator,  to  ob- 

tain the  decimal  denominator, 
100.  Then  the  fraction  is  changed  to  the  decimal  form  by  omitting 
the  dcnomiiuitor.  If  the  intermediate  steps  be  omitted,  the  true 
result  may  be  obtained  as  in  the  Si^cond  operation. 

2.  Reduce  -^  to  its  equivalent  decimal. 

Case  IT  is  AvViat  ?     Give  explanation.     Rule.      Case  HI  is  what? 
Explain  first  operation.     Second. 


4  )  3.00 


REDUCTION.  123 

THIRD  OPERATION.  ANALYSIS.     Dividing  as  in  the  former 

16)  1.0000  example,  -we  obtain  a  quotient  of  3  fig- 

■ 7  urcs,    625.      But    since   we   annexed   4 

.Uu_o,  Jlns.  ciphers,  there  must  be  4  places  in  the 

required   decimal ;   hence  wc  ])refix   1  cipher.     This  is   made  still 
plainer  by  the  following  operation  ;  thus, 

_1     _1_0  00  0     _62  5      —    n(i"^i 

From  these  illustrations  we  derive  the,  following 

Rule.  I.  Annex  ciphers  to  the  numerator,  and  divide  by 
the  denominator. 

II.  Point  off  as  many  decimal  places  in  the  result  as  are 
equal  to  the  number  of  cipJters  annexed. 

Note.  A  common  fraction  can  be  reduced  to  an  exact  decimal  when 
its  lowest  denominator  contains  only  the  prime  factors  2  and  5,  and 
not  otherwise. 

EXAMPLES    FOR   PRACTICE. 

3.  Reduce  f  to  a  decimal. 

4.  Reduce  f  to  a  decimal. 

5.  Reduce  ^|  to  a  decimal. 


Ans. 

.625. 

Ans. 

.9375. 

Ans. 

.08. 

Ans. 

.046875. 

6.  Reduce  |^  to  a  decimal. 

7.  Reduce  ^^  to  a  decimal. 

8.  Reduce  ^-^  to  a  decimal. 

9.  Reduce  f  to  a  decimal. 

10.  Reduce  -^xj  to  a  decimal. 

11.  Reduce  ^g^  to  a  decimal.  Ans.        .00375. 

12.  Reduce  ^J-^  to  a  decimal.  Ans.  .008. 

13.  Reduce  ^  to  a  decimal.  Ans.   .33333-|-. 

NoTF..     The  sign,  -{-,  in  the  answer  indicates  that  there  is  still  a 
remainder. 

14.  Reduce  ^S  to  a  decimal.  Ans.    .513513-f-- 

Note.   The  answers  to  the  last  two  examples  are  called  repeafinr^ 
\decimnh ;  and  the  fi^ixre  3  in  the  13th  example,  and  the  figures  513  in 
Ithe  14th,  are  called  repetends,  because  they  are  repeated,  or  occur  in 
regular  order. 


Third  operation.     Rule,  first  step  ?     Second?     When  can  a  common 
fi'action  be  reduced  to  an  exact  decimal  ? 


124  DECIBIALS. 


ADDITION. 

t,lQ.    1.   What  is  the  sum  of  3.703,  G21.57,  .G72,  and 
20.0U74? 

OPERATION.  Analysis.     "We  -wTlte  the  numbers  so  that  fig- 

3.703  ^^'^■''  °^  ^i^^^  orders  of  units  shall  stand  in  the  same 

r>2i  57  columns;  that  is,  units  under  units,  tenths  under 

p-.^  tenths,  hundredths  under  hundredths,  &c.     This 

'  brings  the   decimal  points   directly  inider   each 

•  other.    Commencing  at  the  right  hand,  we  add 

045.9524:  each  column  separately,  and  cany  as  in  -whole 

numbers,  and  in  the  result  we  place  a  decimal 

point  between  units  and  tenths,  or  directly  under  the  decimal  point 

in  the  numbers  added.     From  this  example  we  derive  the  following 

Rule.     I.    Write   tlie    numbers   so  that   the   dccimql  points 
shall  stand  directly  under  each  other. 

II.  Add  as  in  whole  nwnbers,  and  place   the  decimal  point, 
in  the  result,  directly  under  the  points  in  the  numbers  added. 

/ 

EXAMPLES    FOR    PKACTICE. 

2.  Add         .199  3.  Add 

2.7509 
.25 
.054 


4.015 

0.75 

27.38203 

375.01 

2.5 

415.05703 

Sum,  3.8599 

Amount, 

4.  Add  1152.01,  14.11018,  152348.21,  9.000083. 

Ans.   153523.330203. 

5.  Add  37.03,  0.521,  .9,  1000,  4000.0004. 

Ans.    5038.4514. 

6.  What  is  the  sum  of  twenty-six,  and  twenty-six  Imn- 
dredths ;  seven  tentlis ;  six,  and  eighty-three  tliou.^andths ; 
four,  and  four  thousandths?  Ans.    37.047. 


Explain  the  operation  of  addition  of  decimals.     Give  rule,  first  step. 
Second. 


ADDITION.  125 

7.  "What  is  the  sum  of  thirty-six,  and  fifteen  thousandths  ; 
three  Jiundred,  and  six  hundred  five  ten-tliousandths  ;  five, 
and  three  miUionths  ;  sixty,  and  eighty-seven  ten-millionths? 

Ans.    401.0755117, 

8.  "\7hat  is  the  sum  of  fifty-four,  and  thirty-four  hun- 
dredtlis  ;  one,  and  nine  ten-lhousandtlis  ;  three,  and  two  liun- 
Vlred  seven  millionths ;  twenty-three  tliousandths ;  eight,  and 
nine  tenths;  four,  and  due  hundred  thirty-five  tliousandths? 

Ans.    71.399107. 

9.  ITow  many  yards  in  three  pieces  of  cloth,  tlie  first  piece 
containing  18.375  yards,  the  second  piece  41.G25  yards,  and 
the  third  [)iece  35.5  yards? 

10.  A's  farm  contains  01.843  acres,  B's  contains  143.75 
acres,  C's  218.4375  acres,  and  D's  21.9  acres;  how  many 
acres  in  the  four  farms  ? 

11.  My  farm  consists  of  7  fields,  containing  12 J  acres,  18f 
acres,  9  acres,  24|  acres,  4||  acres,  8f^y  acres,  and  15^^  acres 
respectively  ;  how  many  acres  in  my  farm  ? 

Note.     Reduce  the  common  fractions  to  decimals  before  adding. 

Ans.   93.6375. 

12.  A  grocer  has  2J-  barrels  of  A  sugar,  53.  barrels  of  B 
sugar,  3 1  barrels  of  C  sugar,  3.0 G42  barrels  of  crushed 
sugar,  and  8.925  barrels  of  pulverized  sugar  ;  how  many  bar- 
rels of  sugar  has  he  ?  Ans.    23.8642. 

13.  A  tailor  made  3  suits  of  clothes;  for  the  first  suit  he 
used  2^  }ards  of  broadcloth,  3j\  yards  of  cassiraere,  and  f 
yards  of  satin  ;  for  the  second  suit  2.25  yards  of  broadcloth, 
2.875  yards  of  cassimere,  and  1  yard  of  satin ;  and  for  the 
third  suit  5-^\  yards  of  broadcloth,  and  1^  yards  of  satin. 
ILnv  many  yards  of  each  kind  of  goods  did  he  use?  How 
many  yards  of  all  ?  Ans.  to  last,   18.375. 


126 


DECIMALS. 


SUBTRACTIOX. 


From  91.73  take  2.18.    Analysis.  In  each  of  these 

three  examples,  wo  write  tlie 
subtrahend  under  the  minu- 
end, placing  units  under 
units,  tenths  under  tenths, 
&;c.  Commencing  at  the 
right  hand,  we  subtract  as 
in  whole  numbers,  and  in 
the  remainders  we  place  the 
decimal  points  directly  under 
those  in  the  numbers  above. 
In  the  second  example,  the 
number  of  decimal  places  in 
the  minuend  is  greater  than 
the  number  in  the  subtra- 
hend, and  in  the  tliird  exam- 
ple the  number  is  less.  In 
both  cases,  we  reduce  bouh 
minuend  and  subtrahend  to 
the  same  number  of  decimal 
places,  by  annexing  ciphers; 
or  we  suppose  the  ciphers  to 
be  annexed,  before  performing  the  subtraction.     Hence  the 

RuLK.  I.  Write  the  numbers  so  tliat  the  decimal  points 
shall  staiTil  directly  vnder  each  other. 

11.  Subtract  as  in  tcliole  numbers,  and  place  the  decimal 
point  in  the  result  directly  tinder  the  points  in  the  given  numbers. 

A.  Find  the  {lifFerence  between  714  and  .910.    Ans.  71.3.084. 

f).  How  much  jireater  is  2  than  .298?  Ans.    1.702. 

G.  From  21.004  fake  75  hundredths. 

7.  From  10.0302  take  2  ten-thousandths.      Ans.    10.03. 


153.   1. 

OPER.\TION. 
91.73 

2.18 
Ans.   89.55 

2.  From  2.9185  take  1.42. 

OPERATION. 

2.9185 
1.42 

« 

Ans.    1.4985 

3.  From  124.65  take  95.58746 

OPEEATION. 

124.05 
95.58740 

Alts.   29.0G254 


8.    From  900  take  .009. 


Ans.    899.991. 


9.    From  two  tii<iu-:uid  lake  two  thousandths. 
10.    From  one  take  one  millionth.  Ans. 


.999999. 


Explain  subtraction  of  fractions.     Give  tire  rule,  first  step.     Second. 


MULTIPLICATION.  ^27 

11.  From  four  hundred  twenty-seven  thousandths  take 
four  hundred  twenty-seven  millionths.  Ans.    .426573. 

12.  A  man  owned  thirty-four  hundredths  of  a  townsliip  of 
land,  and  sold  thirty-four  thousandths  of  the  townshij) ;  how 
much  did  he  still  own  ?  J^ns.    .300. 


MULTIPLICATION. 

154.    1.    What  is  the  product  of  .35  multiplied  by  .5  ? 

OPERATION.         Analysis.     ^Ve  perform  the  muhiplication  the 
.35  same  as  in  whole  numbers,  and  the  onl)'  difficulty 

,5  we  meet  with  is  in  pohiting  ofl'  the  decimal  plaices 

.  .  "1  the  product.  To  determine  how  many  places  to 
'"'' point  oft",  we  may  reduce  the  decimals  to  common 
fractions;  thus,  .'60=1^^^  and  .0  =z  -/q.  Perform- 
ing the  multiplication,  and  we  have  -^f^  X  ,\  =  iVoT'  ^^^^  this 
product,  expressed  decimally,  is  .175.  Here  we  see  that  the  prod- 
uct contains  as  many  decimal  places  as  are  contained  in  both  mul- 
tiplicand and  multiplier.     Hence  the  following 

Rule.  Midtiphj  as  in  whole  numbers,  and  from  the  right 
hand  of  the  product  point  off  as  7nany  fgures  for  decimals  as 
there  are  decimal  places  in  both  factors. 

Notes.  1.  If  there  be  not  as  many  figiires  in  the  product  as  there 
are  decimals  m  both  factors,  supply  the  deficiency  by  prefixing  ciphers, 

2.  To  multiply  a  decimal  by  10,"  100,  1000,  &c.,  remove  the  point  as 
many  places  to  the  right  as  there  are  ciphers  on  the  right  of  the  multi- 
plier. 

EXAMPLES. 

2.  Muhiply  1.245  by  .27.  Ans.    .33615. 

3.  Multiply  79.347  by  23.15.  Ans.    1836.88305. 

4.  Multiply  350  by  .7853. 

5.  ]\Iultiply  one  tenth  by  one  tenth.  Ans.    -01, 

6.  Multiply  25  by  twenty-five  hundredths.     Ans.    6.25. 


Explain  multiplication  of  decimals.  Give  ru/e.  If  the  product  have 
less  decimal  places  than  both  factors,  how  proceed  ?  How  multiply  by 
10,  100,  1000,  &e.  ? 


123  DECIMALS. 

7.  Multiply  .132  by  .241.  Ans.  .031812. 

8.  Multiply  24.35  by  10. 

9.  I^Iultiply  .006  by  1000.  Ans.  6. 

10.  Multiply  .23  by  .009.  Ans.    .00207. 

11.  Multiply  sixty-four  thousandths  by  thirteen  milliontha 

Ans.   .000000832. 

12.  Multiply  eighty-seven   ten-thousandths   by  three  hun- 
dred fifty-twc  hundred-thousandths. 

13.  Multiply  one  million  by  one  millionth.  Ans.    1. 

14.  Multiply  sixteen  thousand  by  sixteen  ten-thousandths. 

Ans.    25. G. 

15.  If  a  cord  of  wood  be  worth  2.37  bushels  of  wheat,  how 
many  bushels  of  wheat  must  be  given  for  9.58  cords  of  wood  ? 

Ans.    22.7046  bushels. 


DmSION. 

155 o    1.   What  is  the  quotient  of  .175  divided  by  .5  ? 

OPERATION.       Analysis.     AVe  perfoi-m  the  division  the  same  as 
.5  )  .175        ^^  whole  numbers,  and  the  only  difficulty  we  meet 
— —       with  is  in  pointing  off  the  decimal  places  in  the  quo- 
ins.  cOO  tient.     To  determine  how  many  places  to  point  off, 
we  may  reduce  the  decimals  to  common  fractions;    thus,  .175 zr 
-j^^,  and  .5  z=  -^^.     Performing  the  division,  and  we  have 

175   5   X^$      10       35 
X  — = 


1000   10   1000  $        100 

and  this  quotient,  expressed  decimally,  is  .35.  Here  we  see  that  the 
dividend  contains  as  many  decimal  jdaces  as  are  contained  in  both 
divisor  and  quotient.     Hence  the  following 

Rule.  Divide  as  in  xohole  numhers,  and  from  the  right 
hand  of  the  (piotient  point  off  as  many  places  for  decimals 
as  ike  decimal  places  in  the  dividend  exceed  those  in  the 
divisor. 

Explain  division  of  decimals.    Give  rule. 


DIVISION.  129 

Notes.  1.  If  the  number  of  figures  in  the  quotient  be  less  than  the 
excess  of  the  decimal  phices  in  tlie  dividend  over  those  in  tlie  divisor, 
the  deticiency  must  be  supplied  by  pretixiug  ciphers. 

2.  If  there  be  a  remainder  after  dividing  the  dividend,  annex  ciphers, 
and  continue  the  division :  the  ciphers  annexed  are  decimals  of  the 
dividend. 

3.  The  dividend  must  always  contain  at  least  as  many  decimal  places 
as  the  divisor,  before  commencing  the  division. 

4.  In  most  business  transactions,  tlie  division  is  considered  suffi- 
ciently exact  when  th-e  quotient  is  carried  to  -1  decimal  places,  unless 
great  accuracy  is  required. 

0.  To  divide  by  10,  100,  1000,  &c.,  remove  the  decimal  point  as 
many  places  to  the  left  as  there  are  ciphers  on  the  right  hand  of  the 
divisor. 


EXAMPLES    FOR   PRACTICE. 

2.  Diviele  .G75  by  .15.  Ans.        4.5. 

3.  Divide  .288  by  3.6.  Ans.        .08. 

4.  Divide  81.6  by  2.5.  Ans.   32.04 

5.  Divide  2.3421  by  21.1. 

6.  Divide  2.3421  by  .211. 

7.  Divide  8.207496  by  .153.  Ans.        54.232. 

8.  Divide  12  by  .7854. 

9.  Divide  3  by  3  ;  divide  3  by  .3 ;  3  by  .03  ;  30  by  .03. 

10.  Divide  15.34  by  2.7. 

11.  Divide  .1  by  .7.  Ans.    .142857-f . 

12.  Divide  45.30  by  .015.  Ans.  3020. 

13.  Divide  .003753  by  625.5.  A7is.        .000006. 

14.  Divide  9.  by  450.  Ans.  .02. 

15.  Divide  2.39015  by  .007.  Ans.         341.45. 

16.  Divide  fifteen,  and   eight  hundred  seventy-five   thou' 
sandths,by  twenty-five  ten-thousandths.  Ans.     6350. 

17.  Divide  365  by  100. 

18.  Divide  785.4  by  1000.  Ans.    .7854. 

19.  Divide  one  thousand  by  one  thousandth. 

Ans.    1000000. 


AMien  are  ciphers  prefixed  to  the  quotient  ?  If  there  be  a  remainder, 
how  proceed  ?  If  the  dividend  have  less  decimal  places  than  the  divi- 
sor, how  proceed  ?     How  divide  by  10,    100,  1000,  &c.  ? 

6* 


130  DECIMALS. 

PROMISCUOUS    EXAMPLES. 

1.  Add  six  hundred,  and  twenty-five  thousandths;  four 
tenths ;  seven,  and  sixty-two  ten-thousandths  ;  three,  and  fifty- 
eight  miUionths  ;  ninety -two,  and  seven  hundredths. 

A/is.   702..O01258. 

2.  What  is  the  sum  of  81.003 -f  5000.4' -f  5.0008 + 
73.87563  -f  1000  +  25  +  3.000548  +  .0315  ? 

3.  From  eighty-seven  take  eighty-seven  thousandths. 

4.  What  is  the  difference  between  nine  million  and  nine 
millionths?  Ans.   8099999.999991. 

5.  Multiply  .305  hy  .15.  Ans.   .05475. 

6.  Multiply  three  thousandths  by  four  hundredths. 

7.  If  one  acre  produce  42.57  bushels  of  corn,  how  many 
bushels  will  18.73  acres  produce  ?  Ans.   797.33G1. 

8.  Divide  .125  by  8000.  Ans.    .000015625. 

9.  Divide  .7744  by  .1936. 

10.  Divide  27.1  by  100000.  Ajis.    .000271. 

11.  If  G.35  acres  produce  70.6755  bushels  of  wheat,  what 
does  one  acre  produce  ?  Ans.    11.13  bushels. 

1 2.  Reduce  .625  to  a  common  fraction.  Ans.   |. 

13.  Express  26.875  by  an  integer  and  a  common  fraction. 

Ans.    20 J. 

14.  Reduce  ^f^  to  a  decimal  fraction.  Aiis.  .016. 

15.  Reduce  -~  *°  ^  decimal  fraction.  Ans.   .5. 

16.  How  many  times  will  .5  of  1.75  be  contained  in  .25  of 
171?  •  Ans.   5. 

17.  What  will  be  the  cost  of  3|  bales  of  cloth,  each  bale 
containing  36.75  yards,  at  .85  dollars  per  yard  ? 

18.  Traveling  at  the  rate  of  43.  miles  an  hour,  how  many 
hours  will  a  man  require  to  travel  56.925  niilos. 

Ans.    12§  hours. 


NOTATION   AND   NUMERATION.  131 


DECIMAL   CURRENCY. 

1»"jO.  Coin  is  money  stamped,  and  has  a  given  value  es- 
tablished by  law. 

lejT,  Currency  is  coin,  bank  bills,  treasury  notes,  &c.,  in 
circulntion  as  a  niedimn  of  trade. 

I»)8.  A  Decimal  Currency  is  a  currency  whose  denom- 
inations increase  and  decrease  in  a  tenfold  ratio. 

Note.  The  currency  of  the  United  States  is  decimal  currency,  and 
is  sometimes  called  Federal  Money ;  it  was  adopted  by  Congress  in  178G. 

NOTATION    AND    NUMERATION. 

The  ffold  coins  of  the  United  States  are  the  double  easrle, 
eagle,  half  and  quarter  eagle,  three  dollar  piece,  and  dollar. 

The  silver  coins  are  the  dollar,  half  and  quarter  dollar,  dime 
and  half  dime,  and  three  cent  piece. 

The  nickel  coin  is  the  cent. 

Notes.  1.  The  following  pieces  of  gold  are  in  use,  but  are  not  legal 
coin,  viz. ;  the  fifty  dollar  piece,  and  the  half  and  quarter  dollar  pieces. 

2.  The  copper  cent  and  half  cent,  though  still  in  circulation,  are  no 
longer  coined. 

3.  The  mill  is  used  only  ui  computation ;  it  is  not  a  coin. 

TABLE. 

10  mills  {in.)  make  1  cent,      .  .  .  c. 

10  cents  "       1   dime,    .  .  .  d. 

10  dimes  "       1   dollar,  .  .  .  $. 

10  dollars  "       1  eagle,  .  .  .  E. 

UNIT    EQUIVALENTS. 
Mills.  Cents. 

1*-*  :=    1  BimeH.  •■ 

100       =10       z=  1  !)„„,,, 

1000       =100       =10      =1  E«gle. 

10000  =   1000  =  100  =  10  =  1 

Note.  The  character  .$  is  supposed  to  be  a  contraction  of  U.  S., 
(United  States,)  the  U  being  placed  upon  the  S. 

"What  is  coin  ?  Currency  ?  Decimal  currency  ?  Federal  money  ? 
"What  arc  the  gold  coins  of  U.  S.  ?  Silver  ?  Co])j)er  ?  "What  are  the 
cienominations  of  U.  S.  currency  ?  ^^^lat  js  the  sign  of  dollars  ?  From 
what  derived  ? 


132  DECIMAL   CURRENCY. 

I«>9.  The  dollar  is  the  itnit  of  United  States  money; 
dimes,  cents,  and  mills  are  fractions  of  a  dollar,  and  are  sepa- 
rated from  the  dollar  by  the  decimal  point ;  thus,  two  dollars 
one  dime  two  cents  five  mills,  are  written  $2,125. 

By  examining  the  table,  we  see  that  the  dime  is  a  tenth  part 
of  the  unit,  or  dollar;  the  cent  a  tenth  part  of' the  dime  or  a 
hundredth  part  of  the  dollar;  and  the  villi  a  tenth  part  of  the 
cent,  a  hundredth  part  of  the  dime,  or  a  thousandth  part  of  the 
dollar.  Hence  the  denominations  of  decimal  currency  increase 
and  decrease  the  same  as  decimal  fractions,  and  are  expressed 
according  to  the  same  decimal  system  of  notation ;  and  they 
may  be  added,  subtracted,  multiplied,  and  divided  in  the  same 
manner  as  decimals. 

Dimes  are  not  read  as  dimes,  but  the  two  places  of  dimes 
and  cents  are  appropriated  to  cents ;  thus,  1  dollar  3  dimes 
2  cents,  or  $1.32,  are  read  one  dollar  thirty-two  cents  ;  hence, 

When  the  number  of  cents  is  less  than  10,  we  write  a  cipher 
before  it  in  the  j)lace  of  dimes. 

Note.    The  half  cent  is  frequently  written  as  5  mills  ;  thus,  24 J  cents, 

■written  $.245. 

100.  Business  men  frequently  write  cents  as  common 
fractions  of  a  dollar  ;  thus,  three  dollars  thirteen  cents  are 
Avritten  $3^^^,  and  read,  three  and  thirteen  hundredths  dollars. 
In  business  transactions,  when  the  final  result  of  a  computation 
contains  5  mills  or  more,  they  are  called  one  cent,  and  when 
less  than  5,  they  are  rejected. 

EXAMPLES    FOR   PRACTICE. 

1.  Write  four  dollars  five  cents.  Ans.    $-4.05. 

2.  Write  two  dollars  nine  cents. 

3.  Write  ten  dollars  ton  cents. 

4.  Write  eight  dollars  seven  mills.  Ans.    $8,007. 

^Miat  i>;  the  unit  of  TT.  S.  rnrroncv  ?  AMint  is  the  Eronornl  law  of 
inrreaso  and  drcroa^c  ?  In  pvaotiro,  how  many  decimal  places  are  civen 
to  cents?  In  hnsiness  transactions,  how  are  cents  frequently  written  ? 
"WTiat  is  done  if  the  mills  exceed  5  ?     If  less  than  5  ? 


REDUCTION.  133 

5.  "Write  sixty-four  cents.  Ans.    $0.G4. 

G.  AVrite  three  cents  two  mills. 

7.  AVrite  one  hundred  dollars  one  cent  one  mill. 

8.  Read  $7.93  ;  $8.02  ;  $G.542. 

9.  KeadSo.272;  $100,025;  $17,005. 

10.    Kead  $16,205;  $215,081;  $1000.011;  $4,002. 

REDUCTION. 

116 1 .  By  examining  the  table  of  Decimal  Currency,  we  see 
thai  10  mills  make  one  cent,  and  100  cents,  or  1000  mills, 
make  one  dollar  ;  hence, 

2o  change  dollars  to  cents,  multiply  ly  lOQ ; 'that  is,  annex 
'tivo  ciphers, 

7o  change  dollars  to  mills,  annex  three  ciphers. 
To  change  cents  to  mills,  annex  one  cipher. 

EXAMPLES    FOR   PRACTICE. 

1.  Change  $792  to  cents.  Ans.   79200  cents. 

2.  Change  $36  to  cents. 

3.  Reduce  $5248  to  cents. 

4.  In  G.25  dollars  how  many  cents?         Ans.    G25  cents. 

Note.  To  chanso  dollars  and  cents  to  cents,  or  dollars,  cents,  and 
mills  to  mills,  remove  the  decimal  point  and  the  sign,  $. 

5.  Change  $63,045  to  mills.         Ans.   63045  mills. 

6.  Change  16  cents  to  mills. 

7.  Reduce  S3.008  to  mills. 

8.  In  89  cents  how  many  mills  ?  ■ 

16^.     Conversely, 

To  change  cents  to  dollars,  divide  hy  100  ;  that  is,  point  off 
two  figures  from  the  right. 

To  change  mills  to  dollars,  point  off  three  fgures. 
To  change  mills  to  cents,  point  off'  one  figure. 

How  are  dollars  changed  to  cents  ?  to  mills  ?     How  are  cents  changed 
to  mills?   How  are  cents  changed  to  dollars  ?  MiUs  to  dollars  ?  to  cents? 


131  DECIMAL  CURRENCY. 

teXA-MPLES    FOR    PRACTICE. 

1.  Change  875  cents  to  dollars.  Ans.    $8.75. 

2.  Change  1504  cents  to  dollars. 

3.  In  13875  cents  how  many  dollars? 

4.  In  16525  mills  how  many  dollars? 

5.  Reduce  524  mills  to  cents. 

6.  Reduce  6524  mills  to  dollars. 

ADDITION. 

16!S.  1-  A  man  bought  a  cow  for  21  dollars  50  cents,  a 
horse  for  1 25  dollars  37J-  cents,  a  harness  for  4G  dollars  75  cents, 
and  a  carriage  for  2 1 0  dollars  ;  how  much  did  he  pay  for  all  ? 

OPERATION. 

%     21  50  Analysis.     Writing  dollars  under  dol- 

,^.'   __  lars,  cents  under  cents,  &c.,  so  that  the 

~   '[.-  decimal   points    shall    stand   under    each 

'*  other,  we  add  and  point  off  as  in  addition 

2!  0-00  of  decimals.     Hence  the  following 

Ans.    S403.G25 

Rule.     I.    Write  dollars  under  dollars,  cents  under  cents,  S)-c. 
II.  Add  as  in  simple  numbers,  and  place  the  point  in  the 
amount  as  in  addition  of  decimals. 

examples  for  practice. 

2.  What  is  the  sum  of  50  dollars  7  cents,  1000  dollars  75 
cents,  60  dollars  3  mills,  18  cents  4  mills,  1  dollar  1  cent,  and 
25  dollars  45  cents  8  mills  ?  Ans.   $1137.475. 

3.  Add  304  dollars  54  cents  1  mill,  486  dollars  6  cents,  93 
dollars  9  mills,  1742  dollars  80  cents,  3  dollars  27  cents  G 
mills.  Ans.    $2089.686. 

1.  Add  92  cents,  10  cents  1  mills,  35  cents  7  mills,  18  cents 
G  mills  M  cents  4  mills,  12J-  cents,  and  99  cents.  Ans.  $3,126. 

Explain  the  process  of  addition  of  decimal  currency.  Pvule,  first  step. 
Second. 


SUBTRACTION.  I'J.j 


5.  A  farmer  receives  89  dollars  74  cents  for  wheat,  13  dol- 
lars 3  cents  lor  corn,  G  dollars  374  cents  for  potatoes,  and  11) 
dollars  G2i  cents  for  oats ;  Avliat  does  he  receive  for  the 
Avhole  ?  Jns.    $128.77. 

G.  A  lady  bonght  a  dress  for  9  dollars  17  cents,  trimmings 
for  874  cents,  a  pa[)er  of  i)ins  for  (j^  cents,  some  tai)e  ibr  4 
cents,  some  thread  for  8  cents,  and  a  comb  i'or  11  cents;  what 
did  she  pay  for  all  ?  •  Ans.    $]  0.3375. 

7.  Paid  for  building  a  house  S2175.75,  for  painting  the 
same  $240.37;i,  tor  furniture  $605.40,  for  carpets  $140.12i^; 
what  was  the  cost  of  the  house  and  furnishing  ? 

8.  Bought  a  ton  of  coal  for  S6.08,  a  barrel  of  sugar  for 
$26.G25,  a  box  of  tea  for  §1G,  and  a  barrel  of  flour  for  $7.40  ; 
what  was  I  he  cost  of  all? 

9.  A  merchant  bought  goods  to  the  amonnt  of  $7425.50 ; 
he  paid  for  duties  on  the  same  $253.9 G,  and  for  freight 
$170.09  ;  what  was  the  entire  cost  of  the  good.^  ? 

10.  I  bought  a  hat  for  $3.62^,  a  pair  of  shoes  for  $1£.  an 
umbrella  for  $1|,  a  pair  of  gloves  for  $.G2J,  and  a  cane  for 
$.874  ;  what  was  the  cost  of  all  my  purchases  ?     Ans.   $8.25. 

SUBTRACTIOX. 

IG-l.  1.  A  man,  having  $327.50,  paid  out  $1SG.75  for 
\  horse  ;  Iww  much  had  he  left  ? 

OPERATiox.  Analysis.     Writing  the  less  number  im- 

$327.50  ^^^'  ^^^  greater,  dollars  under  dollars,  cents 

18G  75  under  cents,  Src,  we  subtract  and  ])oiiit  off 

in  the  result  as  in  subtraction  of  decimals. 

Ans.    $140.75  Hence  the  following 

IvULE,  I.  W}-ife  the  siihtrahend  under  the  minuend,  dollars 
under  dollars,  cents  zmder  cents.  S)-c. 

11.  Suhtract  as  in  simple  numbers,  and  place  the  point  in 
the  remainder,  as  in  siibtraction  of  decimals. 

Explain  the  process  of  subtraction.      Give  rule,  first  step.     Second. 


136  DECIMAL    CURRENCY. 

EXAMPLES    FOR    PRACTICE. 

2.  From  $365  dollars  5  mills  take  2G7  dollars  1  cent  8 
mills.  A?is.    $97,987. 

3.  From  50  dollars  take  50  cents.  Ahs.    $49.50. 

4.  From  100  dollars  take  1  mill.  Ans.    $99,999. 

5.  From  1000  dollars  take  3  cents  7  mills. 

6.  A  man  bought  a  farm  for  $1575.24,  and  sold  it  for 
$1834.16;  what  did  he  gain  ?  Ans.    $258.92. 

7.  Sold  a  horse  for  145  dollars  27  cents,  which  is  37  dol- 
lars 69  cents  more  than  he  cost  me  ;  what  did  he  cost  me  ? 

8.  A  merchant  bought  flour  for  $5.62^  a  barrel, and  sold 
it  for  $6.84  a  barrel ;  how  much  did  he  gain  on  a  barrel  ? 

9.  A  gentleman,  having  $14725,  gave  $3560  f(jr  a  store, 
and  $7015.871  for  goods;  how  much  money  had  he  left  ? 

10.  A  lady  bought  a  silk  dress  for  $13|,  a  bonnet  for  $5^,  a 
pair  of  gaiters  for  $1 1,  and  a  fan  for  $| ;  she  paid  to  the  shop- 
keeper a  twenty  dollar  bill  and  a  five  dollar  bill ;  how  much 
change  should  he  return  to  her?  Ans.    $3.75. 

Note.   Heduce  the  fractions  of  a  dollar  to  cents  and  mills. 

11.  A  gentleman  bought  a  pair  of  horses  for  $480,  a  har- 
ness for  $80.50,  and  a  carriage  for  $200  less  than  he  paid  for 
both  horses  and  harness;  what  was  the  cost  of  the  carriage? 

Ans.    $360.50. 

IMULTIPLICATION. 

J  05.  1.  If  a  barrel  of  flour  cost  $6,375,  what  will  85 
barrels  cost  ? 

OrERATION. 

$6,375  Analysis.    We  multiply  as  in  simple 

85  numbers,  always  regarding  the  midtiplicr 

_  as  an  ahsfrncf  number,  and  point  oil'  liom 

the  right  hand  of  the  result,  as  in  multijjli- 

.51000  cation  of  decimals.     Hence  the  following 

Ans.    $541,875 


Give  analysis  for  multiplication  in  decimal  currency. 


DIVISION.  137 

Rule.  Multiply  as  in  sim'ple  numhers,  and  place  the  point 
in  the  product,  as  in  multiplication  of  decimals. 

EXAMPLES    FOR    PRACTICE. 

2.  If  a  cord  of  wood  be  worth  $4,275,  wliat  will  300  cords 
be  worth?  ^"S-    $1282.50. 

3.  What  will  175  barrels  of  apples  cost,  at  $2.45  per  bar- 
rel ?  Ans.    $428.75. 

4.  What  will  800  barrels  of  salt  cost,  at  $1.28  per  barrel? 

5.  A  o-rocer  bought  372  pounds  of  cheese  at  $.15  a  pound, 
434  pounds  of  coffee  at  $.12i  a  pound,  and  16  bushels  of  pota- 
toes at  $.33  a  bushel ;  what  did  the  whole  cost  ? 

6.  A  boy,  being  sent  to  purchase  groceries,  bought  3  pounds 
of  tea  at  56  cents  a  pound,  15  pounds  of  rice  at  7  cents  a 
pound,  27  pounds  of  sugar  at  8  cents  a  pound ;  he  gave  the 
grocer  5  dollars ;  how  ranch  change  ought  he  to  receive  ? 

7.  A  farmer  sold  125  bushels  of  oats  at  $.374-  a  bushel, 
and  received  in  payment  75  pounds  of  sugar  at  $.09  a  pound, 
12  pounds  of  tea  at  $.60  a  pound,  and  the  remainder  in  cash; 
how  much  cash  did  he  receive  ?  Ans.    $32.92.j-. 

8.  A  man  bought  150  acres  of  land  for  $3975  ;  he  after- 
ward sold  80  acres  of  it  at  $32.50  an  acre,  and  the  i-emainder 
at  $34.25  an  acre ;  how  much  did  he  gain  by  the  transaction  ? 

Ans.  $1022.50. 

DIVISION. 

166.    1.   If  125  barrels  of  flour  cost  $850,  how  much 

will  1  barrel  cost  ? 

OPERATION.  Analysis.    We  divide  as  in 

125  )  $850.00  (  $6.80,  Ans.       f™P'e  numbers,  and  as  there 
rr-n  is   a  remainder  after  dividing 


the  dollars,  we  reduce  the  div- 
1000  idend  to  cents,  by  annexinj?  two 

1000  ciphers,  and  continue  the   di- 


0 


vision.     Hence  the  following 


Rule.      Give  rule  for  division  m  decimal  currency. 


138  DECIMAL   CURRENCY. 

Rule.  Divide  as  in  simple  nwnbers,  and  place  the  point  in 
t/ie  quotient^  as  in  division  of  decimals. 

Notes.  1 .  In  bu>iuess  transactions  it  is  never  necessary  to  carry 
the  division  further  than  to  mills  in  tlie  quotient. 

2.  If  the  dividend  will  not  contain  the  divisor  an  exact  nun.ljer  of 
times,  ciphers  may  be  annexed,  and  the  division  continued  as  in  divis- 
ion of  decunals.  In  this  case  it  is  always  safe  to  reduce  the  dividend 
to  mdls,  or  to  3  more  decimal  places  than  the  divisor  contains,  be- 
fore commencing  the  division. 

EXAMPLES    FOR    PRACTICE. 

2.  If  33  gallons  of  oil  cost  $41.25,  what  is  the  cost  per  gal- 
lon ?  Ans.    $1.25. 

3.  If  27  yards  of  broadcloth  cost  $94.50,  what  will  1  yard 
cost? 

4.  If  G4  gallons  of  wine  cost  $136,  what  will  1  gallon  cost? 

Ans.    $2,125. 

5.  At  12  cents  apiece,  how  many  pine-apples  can  be  bouglit 
for  $1.32?  Am.  11. 

G.  If  1  ])0und  of  tea  cost  54  cents,  how  many  pounds  can 
be  bought  for  $405  ? 

7.  If  a  man  earn  $180  in  a  year,  how  much  does  he  earn 
a  month  ? 

8.  If  100  acres  of  land  cost  $2847.50,  what  will  1  acre 
cost?  Ans.   $28,475. 

9.  What  cost  1  pound  of  beef,  if  894  pounds  cost  $80.40? 

Ans.    $.09. 

10.  A  farmer  sells  120  bushels  of  wheat  at  $1,124-  a  bushel, 
for  wliich  he  receives  27  barrels  of  flour;  what  does  the  flour 
cost  him  a  barrel  ? 

11.  A  man  bought  4  yards  of  clotli  at  $3.20  a  yard,  and 
37  pounds  of  sugar  at  $.08  a  pound ;  he  paid  $G.80  in  cash, 
and  ihe  remainder  in  butter  at  $.1G  a  pound  ;  how  many  pounds 
of  Iiiittcr  did  it  take?  Ans.    56  ]iouiids. 

12.  A  man  l)ouglit  an  eqtial  number  of  calves  and  sheep, 
paving  $166.75  for  them  ;  for  the  calves  he  ]>aid  $4.50  ;i  licad, 
and  for  the  sheep  $2.75  a  head  ;  how  many  did  he  buy  of  cacli 
kind?  Ans.   23. 


APPLICATIONS.  139 

13.  If  154  pounds  of  sugar  cost  $18.48,  what  will  1  pound 
cost  ? 

14.  A  merchant  bought  14  boxes  of  tea  for  $500;  it  being 
damaged  he  was  obliged  to  lose  $100.75  on  the  cost  of  it; 
how  much  did  he  receive  a  box  ?  Ans.    $32.37 i-. 

Additional  Applications. 

CASE    I. 

S67.  To  find  the  cost  of  any  number  or  quantit}', 
When  the  price  of  a  unit  is  an  aliquot  part  of  one  dollar. 

168.  An  Aliquot  Part  of  a  number  is  such  a  part  as  will 
exactly  divide  that  number ;  thus,  3,  5,  and  7^  are  aliquot 
parts  of  15. 

Note.     An  aliquot  part  may  be  a  "whole  or  a  niLxcd  number,  while  a 
factor  niu^t  be  a  whole  number.  \ 

'   ALIQUOT    PARTS    OF    ONE    DOLLAR. 

50    cents  =  ^  of  1   dollar.   |   12^  cents  irr   ^   of  1   dollar. 


33^  cents  =  ^  of  1   dollar. 
25     cents  :=  ]-  of  1   dollar. 


4 


20    cents  =  i  of  1  dollar. 
IGg  cents  =  ^  of  1   dollar. 


10     cents  z=z  jify  of  1   dollar. 


81^  cents  =:  jL-  of  1  dollar. 
0^  cents  z=z  -j'g  of  1  dollar. 
5    cents  rr  ^'o  of  1   dollar. 

1.  What  will  be  the  cost  of  3784  yards  of  flannel,  at  25 
cents  a  yard  ? 

OPERATION.  Analysis.      If  the  price  were    $1    a   yard, 

4  ')  378  t  ^^^  ^^^'^  Mould  be  as  many  dollars  as  there  arc 

yards.     But  since  the  price  is  -I-  of  a  dollar  a 

Ans.    $J40  yard,  the  whole  cost  will  be  \  as  many  dollars 

as  there  are  yards  ;  or,  1  of  3784  :=  3784  -^  4  =:  $946.     Hence  the 

Rule.     Take  such  a  fractional  fart  of  the  given  mauler  as 
(he  price  is  part  of  one  dollar. 

EXAMPLES    FOR    PRA'CTICE. 

2.  What  cost  903  bushels  of  oats,  at  33^  cents  per  bushel? 

Ans.    $321. 

Case  I  is  Avhat  ?     AMiat  is  an  aliquot  part  of  a  dollar  ?     Give  ex- 
planation.    Rule, 


140  DECIMAL   CURRENCY. 

3.  AVhat  cost  478  yards  of  delaine,  at  50  cents  per  yard  ? 

4.  AVliat  cost  42GG  yards  of"  sheeting,  at  8^  cents  a  yard? 

Ans.    $355. oO. 

5.  What    cost   1250   bushels   of  apples,  at  12^   cents  per 
bushel?  Ans.    $156.25. 

6.  What  cost  3126  spools  of  thread,  at  6|  cents  per  spool? 

Ans.    $195,375. 

7.  At  IGg  cents  per  dozen,  what  cost  1935  dozen  of  eggs? 

Ans.    322.50. 

8.  What  cost  56480  yards  of  calico,  at  12^  per  yard  ? 

9.  At  20  cents    each   what   will   be   the  cost  of  1275  salt 
barrels?  -^ns.    $255. 

CASE    II. 

169.   The  price  of  one  and  the  quantity  being  given, 
to  find  the  cost. 

1.  How  much  will  9  barrels  of  flour  cost,  at  $6.25  per 
barrel  ? 

OPERATION.  Analysis.     Since  one  barrel  cost  $6.25,9 

$6.25  barrels  will  cost  9  times  $6.25,  and  $6.25  X 

q  9  :=  $56.25.     Hence  j 

Ans.    $56.25 

Rule.     31idtipli/  the  jnice  of  one  by  the  quantity. 

EXAMPLES    FOR    PRACTICE. 

2.  If  a  pound  of  beef  cost  9  cents,  what  will  8G4  pounds 
cost?  Ans.    $77. 7(). 

3.  What  cost  87  acres  of  govornmont  land,  at  $1.25  per 
acre  ? 

4.  What  cost  400  barrels  of  salt,  at  $1.45  per  barrel  ? 

Ans.    $580. 
5     Wliat  cost   TG  chests  of  tea,  each  chest  containing  52 
pounds,  at  44  cents  per  pound? 

Case  II  is  ■\vh;it  ?     Give  explanation.     Rule. 


APPLICATIONS.  141 

CASE   III. 

170 .  The  cost  and  the  quantity  being  given,  to  find 
the  price  of  one. 

1.  If  30  bushels  of  corn  cost  $20.70,  what  will  1  bushel 
cost  ? 

OPERATION.  Analysis.     If  30  bushels  cost  $20.70,   1 

310  )  S'^IO  70  bushel  -will  cost  ^^  of  $20.70;  and  $20.70-^ 

<—^- 30;=  $.69.     Hence, 

$.69 

Rule.     Divide  the  cost  by  the  quantity. 

EXAMPLES    FOR   PRACTICE. 

2.  If  25  acres  of  land  cost  $175,  what  will  1  acre  cost? 

3.  If  48  yards  of  broadcloth  cost  $200,  what  will  1  yard 
cost?  Ans.    $4.1 6§. 

4.  If  96  tons  of  hay  cost  $1200,  what  will  1  ton  cost? 

5.  \'i  10  Unabridged  Dictionaries  cost  $56.25,  what  will  1 
cost  ?  Ans.    $5.62J.. 

6.  Bought  18  pounds  of  tea  for  $11.70;  what  was  the  price 
per  pound  ?  Ans.    $.65. 

7.  If  53  pounds  of  butter  cost  $10.07,  what  will  1  pound 
cost  ? 

8.  A  merchant  bought  800  barrels  of  salt  for  $1016  ;  what 
did  it  cost  him  per  barrel  ? 

9.  If  343  sheep  cost  $874.65,  what  will  1  sheep  cost  ? 

Ans.    '$'2.00. 

10.  If  board  for  a  family  be  $684.37^  for  1  year,  how  much 
is  it  per  day?  Ans.    $1.87^. 

CASE    IV. 

171.  The  price  of  one  and  the  cost  of  a  quantity 
being  given,  to  find  the  quantity. 

1.    At  $6  a  barrel  for  flour,  how  many  barrels  can  be  bought 
for  $840  ? 

Case  III  is  what  ?     Give  explanation.     Rule.     Case  IV  is  what  f 


142  DECIMAL   CURRENCY. 

OPERATION.  Analysis.     Since  .*6  Avill  buy  1  barrel 

Q  \  g4()  of  flour,  $840  Avill  buy  J  as  many  barrels 

as  there  are  dollars,  or  as  manv  barrels  as 

Ans.    140  barrels.         $6  is  contained  times  in  8840 ;  840 -^  6 
rr  140  barrels.     Hence, 

Rule.     Divide  the  cost  of  the  quantity  by  the  price  of  one. 

EXAMPLES    FOU    PRACTICE. 

2.  How  many  dozen  of  eggs  can  be  bought  for  $.5.55,  if  one 
dozen  cost  $.15  ?  Ans.    37  dozen. 

3.  At  $12  a  ton,  how  many  tons  of  hay  can  be  bought  for 
$216?  Ans.    18  tons. 

4.  How  many  bushels  of  wheat  can  be  bought  for  $2178.75, 
if  1  bu.<hel  cost  $1.25?  Ans.    1743  bushels. 

5.  A  dairyman  expends  $G43.50  in  buying  cows  at  $19^ 
apiece  ;  how  many  cows  does  he  buy?  Ans.    33  cows. 

6.  At  $.45  per  gallon,  how  many  gallons  of  molasses  can 
be  bought  for  $52.G5  ? 

7.  A  drover   bought  horses  at   $264  a  pair;  how  many 
horses  did  he  buy  for  $6336? 

8.  At   $65   a  ton,  how  many  tons  of  railroad  iron  can  be 
bought  for  $117715?  Ans.    1811  tons. 

CASE  r. 

172.    To  find  the  cost  of  articles  sold  by  the  100, 
1000,  &c. 

1.    What  cost  475  feet  of  limber,  at  $5.24  per  100  feet  ? 

FIRST   OPERATION. 

$5  24  Analysis.    If  the  ])rice  were  $5.24  per 

._e  foot,  the  cost  of  475  feet  would  bo  475  X 

$5.24  =  .92489.      Rut    since    $5.24    is   the 

2620  price  of  100  feet,  $2480  is  100  times  the  true 

3668  value.     Therefore,  to  obtain  the  true  value, 

2006  ^^'^'  divide  .$2189  by  100,  wliioh  we  ma}'  do 

l)y  cutting  olf  two  fifrurcs  from  the  riiiht,  and 

100  )  $24^S9J)0  t4  result  is  $24.89.     Or, 

Ans.  $24.89 

Give  explanation.     I'lile.     Case  V  is  what  ?     Give  first  explanation. 


APPLICATIONS.  143 

SECOND  OPEKATION.       An'alysis.     Since  1  foot  costs  yl^,  or.  01, 
$5.24  '  '^^  $<5.24,  475  feet  will  cost  *5-|,  or  4. To  times 

4  7.5  $0.24,  M'hich  is  $24.89. 

2620  Note.     For  the  same  reasons,  when  the  price 

np^Q  is  per  thousand,  Ave  divide  the  product  by  1000, 

DODO  pj.^  which  ifs  ttiorc  convenient  in  practice,  we  re- 

2096  duce  the  given  quantity  to  thousands  and  deci- 

nials  of  a  tliousand,  by  pointing  off  three  tigures 

$24.8900  £roni  the  right  hand.     Hence  the 

Rule.  I.  Reduce  the  given  quantity  to  hundreds  and  deci- 
mals of  a  hundred,  or  to  thousands  and  decimals  of  a  thousand. 

II.  Multiply  the  price  by  the  quantity,  and  point  off  in  the 
residt  as  in  midtiplication  of  decimals. 

Note.  The  letter  C  is  used  to  indicate  hundreds,  and  Mto  indicate 
thousands. 

EXAMPLES    FOU    PRACTICE. 

2.  What  will  42650  bricks  cost,  at  $4.50  per  M  ? 

Ans.   $191,925. 

3.  What  is  tlie  freight  on  2489  pounds  from  Boston  to  New 
York,  at  $.85  per  100  pounds  ?  Ans.    $21.1504-. 

4.  What  will  7842   feet  of  pine  boards  cost,  at    $17.25 
j  perM?  Ans.    $135,274+. 

5.  What  cost  2348  pine-apples,  at  $124  per  100  ? 

6.  A  broom  maker  bought  1728  broom-handles,  at  $3  per 
1000  ;  how  much  did  they  cost  liim  ? 

7.  Wiiat  is  the  cost  of  2400  feet  of  boards,  at  $7  perM; 
865  feet  of  scantling,  at  $5.40  per  M;  and  1256  feet  of  lath,  at 
$.80  per  C?  Ans.    $31,519. 

8.  What  will  be  the  cost  of  1476  pounds  of  beef,  at  $4.37^ 
per  hundred  pounds  ? 

CASE    VI. 

373.  To  find  the  cost  of  articles  sold  by  tlic  ton  of 
2000  pounds. 

1.  How  much  will  2376  pounds  of  hay  cost,  at  $9.50  per  ton  ? 
Give  second  explanation.    Rule,  first  step.     Second.    Case  VI  Is  what  ? 


144  DECIMAL   CURRENCY. 

OPERATION.  Analysis.     Since  1  ton,  or  2000  pounds,  cost 

2  )  S9.50  $9.50, 1000  pounds,  or  ^  ton,  will  cost  ^  of  $9.o0, 

— —  or   $9.50  -^2  =  $-1,75.     One   pound   will  cost 

^■^•'^  ToW'  °^  -0^1'  of  $4.75,  and  2376  pounds  will 

2-376  cost  2  3-t6,  or  2.376  times  $4.75,  which  is  $1 1.286. 

$11.28600       ^^^^^' 

Rule.  I.  Divide  the  price  of  1  ton  by  2,  and  the  quotient 
will  be  the  price  of  1000  pounds. 

II.  Midtiply  this  quotient  by  the  given  number  of  pounds 
expressed  as  thousandtlis,  as  in  Case  V. 

EXAMPLES    FOR    PRACTICE. 

2.  At  $7  a  ton,  what  will  1495  pounds  of  hay  cost? 

Ans.    $5.2325. 

3.  At  $8.75  a  ton,  what  cost  325  pounds  of  hay  ? 

Ans.    $1,421+. 

4.  What  is  the  cost  of  3142  pounds  of  plaster,  at  $3.84  per 
ton?  Ans.    $6.032-f. 

5.  What  is  the  cost  of  1848  pounds  of  coal,  at  $5.60  per 
ton?  t 

6.  Bought  125  sacks  of  guano,  each  sack  containing  148 
pounds,  at  $18  a  ton ;  what  was  the  cost? 

7.  What  must  be  paid  for  transporting  31640  pounds  of 
railroad  iron  from  Philadelphia  to  Richmond,  at  $3.05  per 
ton?  Ans.    $48,251. 

Bills. 

174.    A  Bill,  in  business  transactions,  is  a  written  state- 
ment of  articles  bought  or  sold,  together  with  the  prices  of      ; 
each,  and  the  whole  cost.  ' 

Find  the  cost  of  the  several  articles,  and  the  amount  or 
footing  of  the  following  bills. 

Give  explnnntion.     Rule,     What  is  a  bill  ?     Explain  the  manner  of       I 
making  out  a  bill.  ' 


« 


ij 


BILLS. 


145 


Net7  York,  June  20,  1859. 

Bot.  of  Baldwin  &  Sherwood, 
7  yds.  Broadcloth,    (a)  $3.00 


Mr.  John  Rice, 


9 
12 

24 
32 


"     Satinet, 
"     Vesting, 


Cassimere, 
Flannel, 


1.12^ 
.90" 

1.37^ 
.65 


Rec'd  Payment, 


$99,925 
Baldwin  &  Sherwood. 

(2.) 
Daniel  Chapman  &  Co.,  Boston,  Jan.  1,  1860. 

Bo't.  of  Palmer  &  Brother. 
67  pairs  Calf  Boots,  (a)  $3.75 


108 

"    Thick     " 

li 

2.62 

1                        ^^ 

"     Gaiters, 

a 

1.12 

27 

"    Buskins, 

i( 

.86 

[         '             35 

"    Slippers, 

(I 

.70 

50      «    Rubbers, 
[             Rec'd.  Payment, 

a 

1.04  _ 

$717.93 

Palmer  &  Brother, 

By  Geo.  Baker. 

(3.) 


G. 

B. 

Grannis, 

KjH 

A.KLESTON 

.  oep 

Bo't. 

of 

Stewart 

&  I 

325  lbs.  A 

Sugar, 

(a) 

1.07 

148    «   B. 

a 

ii 

.06^ 

286    «    Rice, 

a 

.05 

95    "   0. 

J.  Coffee, 

u 

.12^ 

50  boxes 

Oranges, 

i( 

2.75 

75      « 

Lemons, 

i( 

3.62^ 

12      " 

Raisins, 

li 

2.85  _.. 

$501.75' 
Hec'd.  Payment,  hy  note  at  4  mo. 

jj  p  If  Stewart  &  HAaoiOND. 


146  DECIMAL  CURRENCY. 

(4.) 

,,  ^  p  -r.  St.  Loris,  Oct.  15,  1858. 

Messrs.  Osborn  &  Eaton, 

BoH.  of  Rob't.  H.  Carter  »&  Co., 

20000  feet  Pine  Boards  (o)    %\h     per  M. 

7500    "    Plank,  "  9.50     " 

10750    "    Scantling,        "  6.25     " 

3960    "    Timber,  "         2.62^  " 

5287    "        "  "         3.00     " 


$464.6935 


Rec^d.  Payment, 

Rob't.  H.  Carter  &  Co. 

(5.) 

Mr.  J.  C.  Smith,  Cincinnati,  May  3,  1861. 

Bo't.  of  Silas  Johnson, 


25  lbs. 

Coffee  Sugar, 

fa) 

$.11 

5   " 

Y.  H.  Tea, 

.62^ 

26   " 

Mackerel, 

.06J- 

4  gal. 

Molasses, 

.42 

46  yds. 

Sheeting, 

.09 

30    « 

IBleached  Shirting, 

.14 

6  skeins  Sewing  Silk, 

.04 

4  doz. 

Buttons,                        « 
Silas  Jo 

.12 

I  in  » ,. 

$18,24 

HNSON, 

Per  John  Wise. 

PROMISCUOrS    EXAMPLES. 


I: 


1.  What  will  62.75  tons  of  potash  cost,  at  $124.35  per  ton? 

Ans.    $7802.9625. 

2.  What  cost  15  pounds  of  butter,  at  $.17  a  pound  ? 

Ans.  $2.55.        l|j 

3.  A  cargo  of  corn,  containing  2250  bushels,  was  sold  for    I  ] 
$1406.25  ;  what  did  it  sell  for  per  bushel  ?  Ans.    $^.        |lt| 


PROMISCUOUS  EXAMPLES.  147 

4.  If  12  yards  of  cloth  cost  $48.96,  what  will  one  yard 
cost  ? 

5.  A  traveled  325  miles  by  railroad,  and  C  traveled  .45  of 
that  distance  ;  how  far  did  C  travel  ?       Ans.    146.25  miles. 

6.  If  36.5  bushels  of  corn  grow  on  one  acre,  how  many 
acres  will  produce  657  bushels  ?  Ans.    18  acres. 

7.  Bought  a  horse  for  $105,  a  yoke  of  oxen  for  $125,  4 
cows  at  $35  apiece,  and  sold  them  all  for  $400 ;  how  much 
was  gained  or  lost  in  the  transaction  ? 

8.  A  man  bought  28  tons  of  hay  at  $19  a  ton,  and  sold  it 
at  $15  a  ton  ;  how  much  did  he  lose  ?  Ans.    $112. 

9.  If  a  man  travel  4f  miles  an'  hour,  in  how  many  hours 
can  he  travel  34^  miles  ?  Ans.   7.5  hours. 

10.  At  $.31|-  per  bushel,  how  many  bushels  of  potatoes 
can  be  bought  for  $9  ?  Ans.    28.8  bushels. 

11.  If  a  man's  income  be  $2000  a  year,  and  his  expenses 
$3.50  a  day,  what  will  he  save  at  the  end  of  a  year,  or  365 
days  ? 

12.  A  merchant  deposits  in  a  bank,  at  one  time,  $687.25, 
and  at  another,  $943.64  ;  if  he  draw  out  $875.29,  how  much 
will  remain  in  the  bank  ? 

13.  Bought  288  barrels  of  flour  for  $1728,  and  sold  one 
half  the  quantity  for  the  same  price  I  gave  for  it,  and  the  other 
half  for  $8  per  barrel ;  how  much  did  I  receive  for  the  whole  ? 

Aiis.    $2016. 

14.  What  Avill  eight  hundred  seventy-five  thousandths  of  a 
cord  of  wood  cost,  at  $3.75  per  cord  ?  Ans.    $3.281-j-. 

15.  A  drover  bought  cattle  at  $46.56  per  head,  and  sold 
them  at  $65.42  per  head,  and  thereby  gained  $3526.82  ;  how 
many  cattle  did  he  buy?  A7is.    187. 

16.  If  36.48  yards  of  cloth  cost  $54.72,  what  will  14.25 
yards  cost?  Ans.   $21,375. 

17.  A  house  cost  $3548,  which  is  4  times  as  much  as  the 
furniture  cost ;  what  did  the  furniture  cost  ?        A?is.    $887. 

18.  How  many  bushels  of  onions  at  $.82  per  bushel,  can 
be  bought  for  $112.34? 


148  DECIMAL   CURRENCY. 

19.  If  46  tons  of  iron  cost  $3461.50,  what  will  5  tons  cost? 

20.  A  gentleman  left  his  widow  one  third  of  his  property, 
worth  $24000,  and  the  remainder  was  to  be  divided  equally 
among  5  children  ;  how  much  was  the  portion  of  each  child  ? 

Aiis.    $3200. 

21.  A  man  purchased  one  lot,  containing  160  acres  of  land,  at 
$1.25  per  acre; and  another  lot,  containing  80  acres,  at  $5  per 
acre  ;  he  sold  them  both  at  $2.50  per  acre ;  what  did  he  gain 
or  lose  in  the  transaction  ? 

22.  A  druggist  bought  54  gallons  of  oil  for  $72.90,  and 
lost  6  gallons  of  it  by  leakage.  He  sold  the  remainder  at 
$1.70  per  gallon  ;  how  much  did  he  gain  ?         Ans.    $8.70. 

23.  A  miller  bought  122^-  bushels  of  wheat  of  one  man, 
and  75^  bushels  of  another,  at  $.9of  per  bushel.  He  sold  GO 
bushels  at  a  profit  of  $12.50  ;  if  he  sell  the  remainder  at 
$.81^  per  bushel,  what  will  be  Lis  entire  gain  or  loss  ? 

Ans.    $4.718-f-  loss. 

24.  A  laborer  receives  $1.40  per  day,  and  spends  $.75  for 
his  support ;  how  much  does  he  save  in  a  week  ? 

25.  How  many  pounds  of  butter,  at  $.16  per  pound,  must 
be  given  for  39  yards  of  sheeting,  at  $.08  a  yard  ? 

Ans.    19J-  pounds. 

26.  What  cost  23487  feet  of  hemlock  boards,  at  $4.50  per 
1000  feet?  Ans.    $105.6915. 

27.  A  man  has  an  income  of  $1200  a  year ;  how  much 
must  he  spend  per  day  to  use  it  all  ? 

28.  Bouglit  28  fii-kins  of  butter,  each  containing  56  pounds, 
at  $.17  per  pound  ;  what  was  the  whole  cost  ? 

29.  A  merciiant  bought  16  bales  of  cotton  cloth,  each  bale 
containing  13  pieces,  and  each  piece  26  yards,  at  $.07  per 
yard;  Avhat  did  the  whole  cost  ?  Ans.    $378.56. 

30.  Wiiat  cost  48  68  bricks,  at  $4.75  per  M  ? 

31.  A  farmer  sold  27  bushels  of  potatoes,  at  $.33^  per 
bushel ;  28  bushels  of  oats,  at  $.25  per  bushel ;  and  19  bush- 
els of  corn,  at  $.50  per  bushel ;  what  did  he  receive  for  the 
whole  ?  Ans.    $25.50. 


PROMISCUOUS   EXAMPLES.  X49 

32.  John  runs  32  rods  in  a  minute,  and  Homy  pursues 
him  at  the  rate  of  4t  rods  in  a  minute;  how  long  will  it  take 
Henry  to  overtake  John,  if  John  have  8  minutes  the  start? 

Ans.    21^  minutes. 

33.  If  4f  barrels  of  flour  cost  $32.3,  what  will  7J-  bairels 
cost?  Ans.    ^51. 

3-1.  If  .875  of  a  ton  of  coal  cost  $5,035,  what  will  9^  tons 
cost?  Ans.    $59.57. 

35.  For  the  first  three  years  of  business,  a  trader  gained 
$1200.25  a  year;  for  the  next  three,  he  gained  $1800.62  a 
year,  and  for  the  next  two  he  lost  $950.87  a  year;  supposing 
his  capital  at  the  beginning  of  trade  to  have  been  $5000,  what 
was  he  worth  at  the  end  of  the  eighth  year  ?  Ans.  $12100.87. 

36.  AVhat  will  be  the  cost  of  18G40  feet  of  timber,  at  $4.50 
per  100  ?  Ans.    $838.80. 

2i 

37.  Keduce  777  to  a  decimal  fraction.  Ans.    .78125. 

38.  What  will  1375  pounds  of  potash  cost,  at  $96.40  per 
ton?  Ans.    $06,275. 

39.  Reduce  .5625  to  a  common  fraction.  Ans.    -j%. 

40.  Reduce  /^-^  -^'H^  •'^'^tsj  f)  to  decimals,  and  find  their 
sum.  Ans.    1.464375. 

41.  A  man's  account  at  a  store  stands  thus : . 

Dr.  Cr. 

$4,745  $2.70J- 

2.62^  1.240 

1.27  .624 

.45  3.45" 

5.28A  1.87J 

What  is  due  the  merchant?  Ans.    $4.41^. 

42.  A  gardener  sold,  from  his  garden,  120  bunches  of  on- 
ions at  $.121  a  bunch,  18  bushels  of  potatoes  at  $.62^  per 
bushel,  47  heads  of  cabbage  at  $.07  a  head,  0  dozen  cucum- 
bers at  $.18  a  dozen;  he  expended  $1.50  in  spading,  $1.27 
for  fertilizers,  $1.87  for  seeds,  $2.30  in  planting  and  hoeing; 
what  were  the  profits  of  his  garden  ?  Aiis.    $23.08. 


150  REDUCTION. 

REDUCTION. 

1 73.  A  Compound  Number  is  a  concrete  number  wbose 
value  is  expressed  iu  two  or  more  different  denominations. 

1 TG.  Reduction  is  the  process  of  changing  a  number  from 
one  denomiuatioa  to  anotlier  without  altering  its  value. 

Rc-luction  is  of  two  kinds,  Descending  and  Ascending. 

fi  7 f .    Keduction  Descending  is  cbanging  a  number  of  one 
denomination  to  another  denomination  of  less  unit  value  j  tbus,^ 
61  =  10  dimes  =  100  cents  =  1000  mills. 

1^8.  Reduction  Ascending  is  changing  a  number  of  one 
denomination  to  another  denomination  of  greater  unit  value  ; 
thus,  1000  mills  =  100  cents  =  10  dimes  =  |1. 

IT©.  A  Scale  is  a  series  of  numbers,  descending  or  as- 
cendiiig,  used  in  operations  upon  compound  numbers. 

CURRENCY. 
180.  I.     United  States  Money. 

TABLE. 

10  mills  (m.)  make  1  cent, ct. 

10  cents               "  1  dime, d.  { 

10  dimes              "  1  dollar, $.  ••  { 

10  dollars            "  1  eagle, E.  I 

UNIT   EQUIVALENTS. 

ct.  m. 

d.  1    =.  10 

$  1  =       10  =       100 

E.         1  =     10  =     100  =     1000 
1  =  10  =  100  =  1000  =  10000 
Scale  —  uniformly  10. 

Canada  Money. 
The  currency  of  Canada  is  decimal,  and  the  table  and  denom- 
inations are  the  same  as  those  of  the  United  States  monev. 

Note.  Tlie  decimal  cnrroncy  was  adopted  by  tlio  Canadian  Pailiament 
in  18J8,  a:id  the  Act  took  cffi  cL  in  18;j9.  Previously  the  money  of  Can- 
ada was  reckoned  io  jjounds,  .shillings,  and  pence,  the  same  as  in  EngUuid. 

Coins.     The  silver  coins  are  the  shilling,  or  20-cent  piece, 

the  dime,  and  lialf  dime.     The  copper  coin  is  the  cent. 

Note.  The  20-cent  piece  represents  the  value  of  the  shilling  of  the 
old  Canada  Currency. 


COMPOUND   NUMBERS.  151 

II.     English  Money. 
181.     English  Currency  is  the  currency  of  Great  Britain. 

TABLE. 

4  farthings  (far.  or  qr.)  make  1  penny, d. 

12  pence  "      1  sbilliiig, s, 

20  shillings  .      "      1  pound  or  sovereign,... £,  or  sov. 

UNIT   EQUIVALENTS. 

d.  far. 

s.  I  ==      4 

£,  or  sov.     1  =      12  =     48 
1  =  20  =  240  =  960 
Scale  ascending,  4,  12,  20 ;  descending,  20,  12,  4. 

Note.     1.  Farthings  are  generally  expressed  as  fractions  of  a  penny ; 
thus,  1  far.,  sometimes  called  1  quarter,  (qr.),  =  ^d  ;  3  far.  =  f  d. 

2.  Tlie  gold  coitis  are  the  sovereign  (  =  £1),  and  the  half  sovereign, 
(=  10s.) 

3.  The  silver  CO iTis  are  the  crown  (.=  5s.),  the  half-crown  (=  2s.  Gd.), 
the  shilling,  and  the  six-penny  piece. 

4.  The  copper  coins  are  the  penny,  halfpenny,  and  farthing. 

5.  The  guinea  (=21s.)  and  the  half-guinea  (=10s.   6d.  sterling),  are 
old  gold  coins,  that  are  still  in  circulation,  but  are  no  longer  coined. 

6.  In  France  accounts  are  kept  in  francs  and  decimes.     A  franc  is 
equal  to  18.6  cents  U.  S.  money. 

CASE    I. 

18S.   To  perform  reduction  descending. 
1.  Reduce  2l£  18  s.  10  d.  2  far.  to  fartliings. 

OPERATION.  Analysis.     Since   in  £1   there 

oi4?io      iAi    o^  ^re  20  s.,  in  21  £  there  are  20  s.  x 

zi  Xi  lo  s.  iu  a.  J  lar.  . 

OQ  21  =  420  s.,  and  18  s.  in  the  given 


number  added,  makes  438  s.  in  21  £ 
^^^  *  18  s.     Since  in  1  s.  there  are  12  d., 

■"  ^  in  438  s.  there  are  12  d.  x  438  = 


^2^6  ^-  5256  d.,    and    10  d.    in   the    given 

number  added,   makes   5266  d.  in 

210G6  far.  Ans.  21£  18s.  10 d.     Since  in  1  d.  tiiere 

are  4  far.,  in  5266  d.  there  are  4  far. 
X  5266  =  21064  far.,  and  2  far.  in  the  given  number  added,  makes 
21066  far.  in  the  given  number.     Hence, 


152  REDUCTION. 

PiULE.  I.  Multiply  the  highest  denomination  of  the  given 
number  by  that  number  of  tlie  scale  which  will  reduce  it  to  the 
next  lower  dcnoinination,  and  add  to  the  j^^'oduct  the  given 
number,  if  any,  of  that  lower  denomination,      y 

II.  Proceed  in  the  same  manner  with  the  results  obtained  in 
each  lower  denomination^  until  the  reduction  is  brought  to  the 
denomination  required. 

CASE   11. 

183.    To  perform  reduction  ascending. 

I.  Reduce  21066  fortliings  to  pounds. 

OPERATION.  Analysis.     We  first  divide 

4  V21066  far.  *^^^  '^^^^'^  ^^^-  ^^  ^'  ^^^^'^^^^ 

'  -^^— —  '   '  there  are  \  as  many  pence  as 

12  )  5266  d.+    2  far.  forthings,    and   we    find   that 

2|0  )  43|8  s.  +10  d.  21066  far.  =  52G6  d.  +  a   re- 

91   -P  -i_  1  Q  -  mainder  of  2  far.      We  next 

wl  A  +  1»  3.  ^^_^^  g^gg  _j^  ^^^  j^^  because 

Ans.    21  £  18  s.  10  d.  2  far.  there  are  -J^j  as  many  shillings 

as  pence,  and  we  find  that  5266  d.  =  438  s.  +  10  d.  Lastly  we 
divide  the  438  s.  by  20,  because  there  are  ^V  ^s  many  pounds  as 
shillinf^,  and  we  find  that  438  s.  =  21  £  +  18  s.  The  last  quotient 
with  the  several  remainders  annexed  in  the  order  of  the  succeeding 
denominations,  gives  the  answer  21  £  18  s.  10  d.  2  far.     Hence, 

TiULE.  I.  Divide  the  given  number  bg  that  munbcr  of  the 
scale  which  will  reduce  it  to  the  next  higher  denomination. 

II.  Divide  the  quotient  bg  the  next  higher  number  in  the 

scnle  ;  and   so  2}>'occed   to  the  highest  denomination  required. 

The  last  quotient,  ivith   the  several  remainders  annexed  in  a 

reversed  order,  will  be  the  answer. 

Note.  Eoduction  descending  and  reduction  ascending  mutually 
prove  each  oilier. 

EXAMPLES    FOU    PIIACTICE. 

1.  Ill  14194  farthings  how  many  pounds? 

2.  In  14  £  15  s.  8  d.  2  far.  how  many  farthings  ? 

3.  In  15359  farthings  how  many  pounds? 

4.  In  46  Rov.  12  s.  2  d.  how  many  pence  ? 
6.  In  11186  pence  how  many  sovereigns? 


COMrOUNl)   NUMBERS. 


168 


"WT^IGIITS. 

fiS4.  V/eight  is  a,  measure  of  the  quantity  of  matter  a 
boily  contains,  determined  according  to  some  fixed  standard. 
Tliree  scales  of  weight  are  used  in  the  United  States 
nnd  Great  Britain,  namely,  Troy,  Apothecaries',  and  Avoir- 
dupois. 

I.     Tkoy  Weight. 

I8»>.  Troy  Weight  is  used  in  weighing  gold,  silver,  and 
jewels ;  in  philosophical  experiments,  &c. 

T\BLE. 
24  grains    (gr.)    make  1  pennyweight,,  .pwt.  or  dwt. 

20  jjL'iinyweights      "      1   ounce, oz. 

12  ounces  "      1  pound, lb. 

UNIT    EQUIVALENTS. 

invt.  pr. 

1  =       24 

lb.  1  =     20  =     480 

1  =  12  =  240  =  5760 

Scale  — ascending,  21,  20,  12;  descending,  12,  20,  24. 

EXAMPLES    FOR    TRACTICE. 

2.    How    many   pounds    in 
85894  grains  ? 

OPEEATION. 

24  )  85894  gr. 


1.  How  many  grains  in 
10  oz.  18  pwt.  22  gr.? 

141b. 

operatic:^. 

141b.  10  oz.  18  pwt. 
12 

22  gr. 

178  oz. 
20 

3578  pwt. 
24 

14334 
715G 

20  )  3578  pwt.  4-  22  gr. 
12)  178  oz.  4-  18  pwt. 
141b.  +  10oz. 

Ans.    141b.  10  oz.  18  pwt. 


22  gr. 


85894  gr.,  Ans. 

3.  In  5  11).  7  oz.  12  pwt.  9  gr.,  how  many  grains? 

4.  In  32457  grains  how  many  pounds  ? 

Define  weight.     Troy  weight.     Rejoeat  the  table.     Give  the  scale, 
7* 


154 


EEDUCTION. 


5.    Reduce  41 7  GO  grains  to  pounds.         Ans.    7  lb.  3  oz. 
G.    A  miner  had   14  lb.   10  oz.  18  pwt.  of  gold  dust;  how 
much  was  it  worth  at  $.75  a  pwt.  ?  Ans.    $2683.50. 

7.  How  many  spoons,  each  weighing  2  oz.  15  pwt.,  can  be 
made  from  5  lb.  G  oz.  of  silver  ?  Ans.    24. 

8.  A  goldsmith  manufactured  1  lb.  1  pwt.  IG  grs.  of  gold  into 
rings,  each  weigliing  4  pwt.  20  gr. ;  he  sold  the  rings  for  $1.25 
apiece  ;  how  much  did  he  receive  for  them  ?     Ans.    $62.50. 

II.     Apothecaries'  Weight. 
186.    Apothecaries'  WeigM  is  used  by  apothecaries  and 
physicians    in    compounding    medicines ;    but   medicines   are 
bought  and  sold  by  avoirdupois  weight. 

TABLE. 
20  grains  (gr.)  make  1  scruple, sc.  or  9- 


3  scruples 

8  drams 

12  ounces 


1  dram, dr.  or   3  ■ 

1   ounce, oz.  or   §  . 

1  pound, lb.  or  tb. 


lb. 


UKIT    EaUIVALENTS. 

SC.  gr. 

dr.  1    =  20 

1  =      3  —  60 

1  =     8  =     24  =  480 


1  :=  12  —  96  =  288  z=  5760 
Scale  —  ascending,  20,  3,  8,  12;  descending,  12,  8,  3,  20. 

EXAMPLES    FOR    PRACTICE. 


1.    IIow  many  gr.  in  12  lb 
8§  33  1  9  15  gr.? 

OPEIIATION. 

12  ft  85  3  3  19  I5gr. 
12 


152  § 

8 


1219  5 
3 


3658  3 
20 


gr 


2.    How  many  lb  in  73175 

OPERATION. 

2|0  )  7317|5gr. 
3)  3G58  9-f  15gr. 
8  )  1219  5  +  1  9 
12)  152g-{-3  3 
12tb-f-8S 

Ans.  121b  8S  3  3  19  I5gr. 


73175  gr.,  Ans. 
Define  apothecaries'  weight.     Repeat  tlie  table.     Give  the  scale. 


COMPOUND   NUMBERS.  155 

3.  In  IG  lb.  11  oz.  7  dr.  2  sc.  19  gr.,  how  many  grains? 

4.  Reduce  47  ft  6  §  4  3  to  scruples.         A)is.    13092  sc. 

5.  How  many  pounds  of  medicine  would  a  physician  use  in 
one  year,  or  365  days,  if  he  averaged  daily  5  prescriptions 
of  20  grains  each  ?  Ans.    G  ft.  4  §  1  9. 

III.     Avoirdupois  Weight. 
1S7.    Avoirdupois  Weight  is  used  for  all  the  ordinary  pur- 
poses of  weighing. 

TABLE. 

16  drams  (dr.)  make  1  ounce, oz. 

16  ounces  "  1  pound, lb. 

100  lb.  "  1  hundred  weight,  .  cwt. 

20  cwt.,  or  2000  lbs.,      "  1  ton, T. 

UNIT    ZaUIVALENTS. 

OZ.  dr. 

lb.        1  =  16 

cwt.      1  =:    16  =r    2j6 

T.     1  rr:  100  =  1600  —  25600 

1  =:  20  =  2000  =  32000  =  512000 

Scale— ascending,  16,  16,  100,  20  ;  descending,  20,  100,  16,  16. 

Note.  The  Imig  or  grross  ton,  hundred  weight,  and  quarter  were 
formerly  in  common  vise ;  hut  they  are  now  seldom  used  except  in 
estimating  English  goods  at  the  U.  S.  custom-houses,  and  in  freighting 
and  wholesaling  coal  from  the  Pennsylvania  muxes. 

LONG   TON    TABLE. 

28  lb.  make  1  quarter,  marked     qr. 

4  qr.  =  112  lb.  "       1  hundred  weight,        "         cwt. 

20  cwt.  =  2240  lb.         "      1  ton,  "  T. 

Scale  —  ascending,  28,  4,  20;  descending,  20,  4,  28. 

The  following  denominations  are  also  in  use.  • 

56  pounds  make  1  firkin  of  butter. 

100        "  "  1  quintal  of  dried-  salt  fish. 

100        "  "  1  cask  of  raisins. 

196        "  "  1  barrel  of  flour. 

200        "  "  1       "       "   beef,  pork,  or  fish. 

280        "  "  1       "       "  salt  at  the  N.  Y.    State  salt  works. 

56        "  "  1  bushel "     "         "  "  "         "        " 

32        "  "  1       "      "   oats. 

48        «  "  1       '<       «   barley. 

56        "  "  1       "       *'   corn  or  rye. 

60        "  "  1       «       "   wheat. 

Define  avoirdupois  weight.  Repeat  the  table.  Give  the  scale.  The 
long  ton  table.  What  other  denominations  are  in  use  J  What  is  the 
value  of  each  ? 


156  REDUCTION. 


EXAMPLES  FOR  PRACTICE. 


1.   In  25  T.  15  cwt.  70  lb. 
how  many  pounds  ? 

OPERATION. 

25  T.  15  cwt.  70  lb. 

20 


515  cwt. 
100 


2.    In   51570   pounds  how 

many  tons  ? 

OPERATION. 

100)  515701b. 

210)5115  cwt. +  70  lb. 
25T.-[-15cwt. 

Ans.   25  T.  15  cwt.  70  lb. 
515701b.,  Ans. 

3.  Keduee  3  T.  14  cwt.  74  lb.  12  oz.  15  dr.  to  drams. 

4.  Reduce  1913551  drams  to  tons. 

5.  A  tobacconist  bought  3  T.  15  cwt.  20  lb.  of  tobacco,  at 
22  cents  a  pound;  liow  much  did  it  cost  him  ?  Ans.  $1G54.40. 

6.  How  much  will  115  pounds  of  hay  cost,  at  $10  per  ton  ? 

7.  A  grocer  bought    10   barrels  of  sugar,   each  weighing 

2  cwt.  17  1b.,  at  6  cents   a  pound;  5  barrels,  each  weighing 

3  cwt.  G  lb.,  at  7^  cents  a  pound  ;  he  sold  the  whole  at  an 
average  price  of  8  cents  a  pound  ;  how  much  was  his  whole 
gain?  J^ns.    $51.05. 

8.  Paid  $360  for  2  tons  of  cheese,  and  retailed  it  for  12^ 
cents  a  pound  ;  how  much  was  my  whole  gain  ?  Ajis.   $140. 

9.  If  a  person  buy  10  T.  6  cwt.  3  qr.  14  lb.  of  English  iron, 
by  the  long  ton  weight,  at  6  cents  a  pound,  and  sell  the  same 
at  $130  per  short  ton,  how  much  will  he  gain  ?      Ans.    $115.85. 

10.  A  farmer  sold  2  loads  of  corn,  weighing  2352  lbs.  each, 
at  $.90  per  bu. ;  what  did  he  receive  ?  A7is.  $75.60. 

1 1.  How  many  pounds  in  300  barrels  of  flour  ? 

12.  A  grocer  bought  3  barrels  of  salt  at  $1.25  per  barrel, 
and  retailed  it  at  £  of  a  cent  per  pound  ?  what  did  he  gain  ? 

Ans.    $2.55, 

STANDARD    OF    WEIGnT. 

188.  In  the  year  1834  the  U.  S.  government  adopted  a 
uniform  standard  of  weights  and  measures,  for  the  use  of  the 
custom  houses,  and  the  otherbranches  of  business  connected  with 
the  general  government.  Most  of  the  States  which  have  adopt- 
ed any  standards  have  taken  those  of  the  genci'al  government. 


COMPOUND   NUMBERS.  157 

IS??.  Tlie  United  States  standard  unit  of  weight  is  tlie 
Troy  pound  of  the  mint,  wliich  is  tlie  ?ame  as  the  imperial 
standard  pound  of  Great  Britain,  and  is  determined  as  fol- 
lows :  A  cuhic  inch  of  distilled  water  in  a  vacuum,  weighed 
by  brass  weights,  also  in  a  vacuum,  at  a  temperature  of  62° 
Fahrenheit's  thermometer,  is  equal  to  252.458  grains,  of  v/hich 
the  standard  Troy  pound  contains  57G0. 

I!I4>.  I'he  U.  S.  Avoirdupois  pound  is  determined  from 
tlie  standard  Troy  pound,  and  contains  7000  Troy  grains. 
Hence,  the  Troy  pound  is  ^l%%  =  -fft  ^^  ^^  avoirdupois 
pound.  But  the  Troy  ounce  contains  -ff-  ==  480  grains,  and 
the  avoirdupois  ounce  ^f  §^  =  437.5  grains  ;  and  an  ounce  Troy 
is  480 — 437.5  =1  42.5  grains  greater  than  an  ounce  avoirdu- 
po'S.  The  pound,  ounce,  and  grain,  Apothecaries'  weight, 
are  the  same  as  the  like  denominations  in  Troy  weight,  the  only 
diflerence  in  the  two  tables  being  in  the  divisions  of  the  ounce. 

11^1.  COMPAKATIVE   TABLE   OF   WEIGHTS. 

Troy.  Apothecaries'.  Avoirdupois. 

1  pound  z=  5760  grains,  z=z  5760  grains,  rr:  7000  grains. 
1  ounce  rr:    480        "        r=    480        "        z=  437.5      " 

175  i^ounds,  zir    175  pounds,  z=.  144  pounds. 

EXAMPLES    FOB   PRACTICE. 

1.  An  apothecary  bought  5  lb.  10  oz.  of  rhubarb,  by 
avoirdupois  weight,  at  50  cents  an  ounce,  and  retailed  it  at 
12  cents  a  di'am  apothecaries'  weight ;  how  much  did  he  gain  ? 

Ans.    $33.75. 

2.  Change  424  drams  apothecaries'  weight  to  Troy  weight. 

Ans.    4  lb.  5  oz. 

3.  Change  20  lb.  8  oz.  12  pwt.  Troy  weight  to  avoirdu- 
pois weight.  Ans.    H^j-b  ^'^* 

4.  Bought  by  avoirdupois  weight  20  lb.  of  opium,  at  40 
cents  an  ounce,  and  sold  the  same  by  Troy  Aveight  at  50  cents 
an  ounce;  how  much  was  gained  or  lost?        Ans.  $17.83J. 

^^^lat  is  the  TJ.  S.  standard  of  \veio;ht  ?  How  obtained  ?  How  is 
the  avoirdupois  pound  determined  ?  How  is  the  apothecaries'  pound 
determined  ?  What  are  the  values  of  the  denominations  of  Troy,  avoii'- 
dupois,  and  apothecaries'  weight  f 


158  REDUCTION. 


MEASURES   OF  EXTENSION. 

fi3*2.    Extension  has  three  dimensions  —  length,  breadth, 
and  thickness. 

A  Line  has  only  one  dimension  —  length. 

A  Surface  or  Area  has  two  dimensions  —  length  and  breadth. 

A  Solid  or  Body  has  three  dimensions  —  length,  breadth,  and 
thickness. 

I.    Long  Measure. 

193.    Long  Measure,  also  called  Linear  Measure,  is  used 
iu  measuring  lines  or  distances. 

TABLE. 

12     inches  (in.)  make  1  foot, ft. 

3     feet  "  1  yard, yd. 

5^  yd.,  or  16^  ft.,  "  1  rod rd. 

40     rods  "  1  furlong, fur. 

8     furlongs,  or  320  rd.,     "  1  statute  mile,,  .mi. 


UNIT    EQUIVALENTS. 

ft.  in. 

yd.                 1  =:  12 

rd.             "l     =         3  =  36 

f„r.            1  =          5i  rr        16^  =  198 

„,i         1  zr:     40  r=     220     =     6G0  =:  7920 

1  =z  8  =:  320  =  1760     zr:  5280  =  63360 

Scale  —  ascending,  12,  3,  5|,  40,  8  ;  descending,  8, 40,  o^,  3,  12. 
The  following  denominations  are  also  in  use:  — 

3  barleycorns    make    1    i^eh,  ?  "'^^  ^^f^7™^V'V"  "'^'"""° 

•'  '  }  the  length  of  the  foot. 

4  inches  «       1  hand  ^  "^^'^  ^"   measuring   the   height  of 

'  }  horses  directly  over  the  fore  feet. 
6  feet  "      1  fathom,  used  in  measuring  dc])ths  at  sea.  , 

1  1  j;  <,f^f„f„  .v,;i„     u       1  !-•         •!     S  nsed  in  measuring  dis- 

1.15  statute  mues  "      1  geographic  mile,  <  ,  ^  ° 

'^    o    I  '  ^  tances  at  sea. 

3  geographic    "       "       1  league. 

60         "  "       "   ^  1   d  o-  •   .  5  "^  latitude   on  a  meridian  or  of 

69.16  statute   "       "    ^  ^  oegree  j  ]^^„^^^^q  p,^  ^j^g  equator. 

u  60  degrees  "      the  circumference  of  the  earth. 

How  mnnv  flimnnsioiis  has  extension  ?  Define  a  line.  Surface  or 
Rvea.  A  solid  or  body.  Define  lonj];  measure.  \Miat  arc  the  denom- 
inations ?     The  value  of  eaoh.     "What  other  denominations  arc  used  ? 


COMPOUND   NUMBERS. 


159 


Notes.  1.  For  the  purpose  of  measuring  cloth  and  other  goods  sold 
by  the  yard,  the  yard  is  divided  into  halves,  fourths,  eighths,  and  fcLx.- 
teenths.     The  old  table  of  cloth  measure  is  practically  obsolete. 

2.  The  geographic  mile  is  ^^g-  of-j^g  or  -jieg-g  of  the  distance  round 
the  center  of  the  earth.  It  is  a  small  fraction  more  than  1.15  statute 
miles. 

3.  The  length  of  a  degree  of  latitude  varies,  being  68.72  miles  at  the 
equator,  68.9  to  69.05  miles  in  middle  latitudes,  and  69.30  to  69.31  miles 
in  the  polar  regions.  The  mean  or  average  length  is  as  stated  in  the 
table.  A  degree  of  longitude  is  greatest  at  the  equator,  where  it  is 
69.16  miles,  and  it  gradually  decreases  toward  the  poles,  where  it  is  0. 


EXAMPLES    FOR    PRACTICE. 


1.    In  2  mi.  4  fur.  32  rd. 
2  yd.  how  many  inches  ? 

OPERATION. 

2  mi.  4  fur.  32  rd.  2  yd. 

8 


20  fur. 
40 


832  rd. 
5J- 


416 
4162 

4578  yd. 
3 


13734  ft. 
12 


2.    In   164808   inches   Low 
many  miles  ? 

OPERATION. 

12)  164808  in. 
3)  13734  ft. 

54  ]  4578  yd. 

2"  J'       2 

11  )9156 

4|0)83|2rd.  + 1-  yd.  =  2  yd. 
8  )  20  fur.  -f  32  rd. 
2  mi.  -|-  4  fur. 
Ans.    2  mi.  4  fur.  32  rd.  2  yd. 


164808  in.,  J«s. 

3.  The  diameter  of  the  earth  being  7912  miles,  how  many 
inches  is  it?  Ans.    501304320  inches. 

4.  In  168474  feet  how  many  miles  ? 

5.  In  31  mi.  7  fur.  10  rd.  3  yd.,  how  many  feet  ? 

6.  If  the  greatest  depth  of  the  Atlantic  telegraphic  cable 
from  Newfoundland  to  Ireland  be  2500  fathoms,  how  many 
miles  is  it  ?  Ans.    2  mi.  6  fur.  29  rd.  1^^  ft. 


160  .  REDUCTION. 

7.  If  this  cable  be  2200  miles  in  length,  and  co.-t  10  cents 
a  foot,  what  -was  its  whole  cost?  Ans.   SllGlGOO. 

8.  A  pond  of  water  measures  4  fathoms  3  feet  8  inclie?  in 
depth  ;  how  many  inches  deep  is  it  ?  Ans.    232. 

9.  How  many  times  will  the  driving  wheels  of  a  locomo- 
tive turn  round  in  going  from  Albany  to  Boston,  a  distance  of 
200  miles,  supposing  the  wheels  to  be  18  ft.  4  inches  in  cir- 
cumference? Ans.    57C00  times. 

10.  If  a  vessel  sail  120  leagues  in  a  day,  how  many  stat- 
ute miles  does  she  sail  ?  Ans.    414. 

11.  How  many  inches  high  is  a  horse  that  measures  14^ 
hands?  Ans.    58. 

surveyors'  long  measure. 

134.  A  Gunter's  Chain,  used  by  land  surveyors,  is  4  rods 
or  6G  feet  long,  and  consists  of  100  links. 

TABLE. 

7.92  inches  '       (in.)  make  1  link, 1. 

25  links  "  1  rod, rd. 

4  rods,  or  66  feet,     "  1  chain   .  .ch.  : 

80  chains  "  1  mile,  . .  mi. 

UNIT    EQUIVALEXTS. 

1.         iP. 

H.     1  =     7.92 

,,,      1  r=   25  =   198 

.     i"—   4  =  100  z=   792 

1*=  80  =  320  =  8000  =r  63360 

Scale— ascending,  7.92,  25,  4,  80  ;  descending,  80,  4,  25,  7.92. 

Note.  Rods  are  seldom  used  in  chain  measure,  distances  being 
taken  in  chains  and  links.  , 

EXAMPLES    FOR    PRACTICE.  W^ 

1.  In  3  mi.  51  ch.  73  1.  how  many  links?  » 

2.  Reduce  29173  I  to  miles. 

3.  A  certain  field,  enclosed  by  a  board  fence,  is  17  ch.  311. 
long,  and  12  ch.  87  1.  wide  ;  how  many  feet  long  is  the  fence 
whicli  encloses  it?  ^ns.    3983.7G  ft. 


Repeat  the  table  of  surveyors'  long  measure.     Give  the  scale. 


COMPOUND   NUMBERS. 


1.61 


II.     Square  Measure. 

S9*9.    A  Square  is  a  figure  having  four  equal   sides,  and 
four  equal  angles  or  corners. 

1  square  foot  is  a  figure  having  four 
sides  of  1  ft.  or  12  in.  each,  as  shown 
in  the  diagram.  Its  contents  are  12 
X  12  1=  144  square  inches.     Hence 

The  contents  or  area  of  a  square,  or 
of  any  other  figure  having  a  uniform 
length  and  a  uniform  breadth,  is  found 
by  multiplying  the  length  by  the  breadth. 
Thus,  a  square- foot  is  12  in.  long  and  12  in.  wide,  and  the  con- 
tents are  12  X  12  =  144  square  inches.  A  board  20  in.  long 
and  10  in.  wide,  is  a  rectangle,  containing  20  X  10  =  2U0 
square  inches. 

1J>6 .    Square  Measure  is  used  in  computing  areas  or  sur- 
faces ;  as  of  land,  boards,  painting,  plastering,  paving,  &c. 

TABLE. 


1 

1 

n 

= 

= 

1 

ft 

12  iu. 


:1  ft. 


144    square  inches  (sq.  iu.)  make  I  square  foot,  marked  sq.  ft. 

9    square  feet  "       1  square  yard,       "      sq.  yd. 

30-^  square  yards  "       1  square  rod,         "      sq.  rd. 

40    square  rods  .  "       1  rood,  "  E,. 

4    roods  "       1  acre,  "  A. 

640    acres  "       1  square  mile,       "      sq.  mi. 


UNIT    EQUIVALENTS 


R. 

1: 
4: 


A. 
80.  mi.      1  ;^ 

1  =  640  =  2o60  =  102400  =  3007600 


Bq.rd. 

40  r= 
160  = 


pq.  yd. 

1  : 

30^-: 

1210: 

4S40: 


sq.  ft. 
1 

9 

2721 : 

10S90 : 

43r)00  : 

: 27878400 : 


sq. in. 

144 

1296 

39204 

1568160 

6272640 

4014489600 


Scale  — ascending,  144,  9,  301  40,  4,  640;  descending,  640,  4, 
40,  301  9,  144. 


Define  a  square.  How  is  the  area  of  a  square  or  any  rectanprular 
fiijnre  found  ?  For  what  is  square  measure  used  ?  Repeat  the  table. 
Give  the  scale. 


162  REDUCTION. 

Artificers  estimate  their  work  as  follows : 

By  the  square  foot :  glazing  and  stone-cutting. 

By  the  square  yard  :  painting,  plastering,  paving,  ceiling,  and 
paper-hanging. 

By  the  square  of  100  feet:  flooring,  partitioning,  roofing,  slating, 
and  tiling. 

Brick-laying  is  estimated  by  the  thousand  bricks;  also  by  the 
square  yard,  and  the  square  of  100  feet. 

Notes.  1.  In  estimating  the  painting  of  moldings,  cornices,  &c.,  the 
measuring-line  is  carried  into  all  the  moldings  and  cornices. 

2.  In  estimatmg  brick-laying  by  the  square  yard  or  the  square  oi 
100  feet,  the  work  is  understood  to  be  11  bricks,  or  12  inches,  thick. 

EXAJIPLES    FOR    PRACTICE. 

1.  In  10  A.  1  R.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 
now  many  square  inches  ? 

OPERATIOX. 

10  A.  1 R.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 
4 

41  R. 

40 


1665  sq.  rd. 


301 


416^ 
49966 

503821  sq.  yd. 
9 


4534441  sq.ft.  ■! 

144  ^ 


36  =  I  sq.  ft.  £ 

1813912  with  136  sq.  in.  '^ 

1813776 

453444 

65296108  sq.  in.,  Ans. 
2.    In  65296108  sq.  in.  how  many  acres  ? 

IIow  do  artisans  estimate  work  ? 


COMPOUND   NUMBERS.  163 


OPERATION. 

144  )  6o29Gl()8  sg.  in. 

9  )  4  j344o  sq.  ft.  -f  28  sq.  in. 

30f  I  50382  sq.  yd.  +  7  sq.  ft. 
4  4 


121  )  201  r>28  fourths  sq.  yd. 

4.0)166|5  sq.  rd.  -\-\--  =  15|  sq.  yd. 
4)4111. -[-25  sq.rd. 

10  A.  -(-  1  R. 

Ans.    10  A.  1  11.  25  sq.  rd.  15f  sq.  yd.  7  sq.  ft.  28  sq.  in. 

f  10  A.  1  II.  25  sq.  rd.  15  sq.  yd,  7  sq.  ft.  28  sq.  in. 

Or  j  C  sq.  ft.  108  sq.  in. 

Or       10  A.  1  II.  25  sq.  rd.  IG  sq.  yd.  x  sq.  ft.  136  sq.  i 


in. 


Analysis.  Dividing  by  the  numbers  in  the  ascending  scale,  and 
arranging  the  remainders  according  to  their  order  in  a  line  below, 
we  find  the  square  yards  a  mi.xed  number,  15|.  Ijut  ^  cf  a  sq.  yd. 
=  I  of  9  sq.  ft.  =:  6|  sq.  ft.  ;  and  |  of  a  sq.  ft.  =i  |  of  144  sq.  in.  rr 
108  sq.  in.  Therefore  |  sq.  yd.  =:r  6  sq.  ft.  108  sq.  in.  ;  and  adding 
108  sq. in.  to  28  sq.  in.  we  have  136  sq.  in.,  and  G  sq.  ft.  to  7  sq.ft.  we 
have  13  sq.  ft.  =  1  sq.  yd.  4  sq.  ft.,  and  writing  the  4  sq.  ft.  in  the 
result,  and  adding  1  sq.  yd.  to  15  sq.  yd.  we  have  for  the  reduced 
result,  10  A.  1  11.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 

?,.  Reduce  87  A.  2  R.  38  sq.  rd.  7  sq.  yd.  1  sq.  ft.  100  sq. 
in.  to  square  inches.  ^ns.    55035oOG8  sq.  in. 

4.  Reduce  550355068  square  inches  to  acres. 

5.  A  field  100  rods  long  and  30  rods  wide  contains  how 
many  acres  ?  Ans.    18  A.  o  R. 

G.  How  many  rods  of  fence  will  enclose  a  faim  a  mile 
square?  -4?«s.    1280  rods. 

7.  How  much  additional  fence  will  divide  it  into  four  equal 
square  fields  ?  ^"S-    640  rods. 

8.  TTow  many  acres  of  land  in  Boston,  at  $1  a  square  foot, 
will  $100000  purchase? 

Ans.    2  A.  1  R.  7   sq.  rd.  9  sq.  yd.  3^  sq.  ft. 

9.  How  many  yards  of  carpetinjr,  1  yd.  wide,  will  be  required 
to  carpet  a  room  18^  ft.  long  and  16  ft.  wide  ?  Ans.  32|  yd. 


164:  REDUCTION. 

10.  "What  would  be  the  cost  of  plastering  a  room  18  ft.  long, 
1G4-  ft.  wide,  and  9  ft.  high,  at  22  cts.  a  sq.  yd.?  Ans.  $22.44. 

11.  What  will  be  the  expense  of  slating  a  roof  40  feet 
long  and  each  of  the  two  sides  20  feet  wide,  at  $10  per 
square?  ^«s-    Si  GO. 

surveyors'  square  measure. 
1397.    This  measure  is  used  by  surveyors  in  computing  the 
area  or  contents  of  land. 

TABLE. 

625  square  links  (sq.  1.)  make  1  pole, P. 

1(5  poles  "      1  square  chain,,  .sq.  ch. 

10  square  chains  "      1  acre V. 

6-iO  acres  "      1  square  mile,.  ..sq.  ini. 

36  square  miles  (6  miles  square)     "      1  township,. ..    .  ...Tp. 

LNIT    r.aUIVALEXTS. 

r.  Fq.  1. 

sq.  ch.  1  ^^^^  "-'^ 

A        1  =     16  =      1000 

gn  mi      l'=:     10  —     160  =  100000 

Tp    1  =   610  =r   6400  =    102400  =      64000000 

1  "i=  36  =  23040  —  231)400  =  368G400  =  2304000000 

Scale  — ascending,  625,  16,  10,  640,  36;  descending,  36,  640, 

10,  16,  623. 

Notes.     1.  A  sqiiare  mile  of  land  is  also  called  a  section. 

2.  Canal  and  railroad  engineers  commonly  use  an  engineers'  chain, 
■which  consists  of  100  links,  each  1  foot  long. 

3.  The  contents  of  land  are  commonly  estimated  in  square  miles, 
acres,  and  hundredths  ;  the  douommation,  rood,  is  fast  going  mto  dis- 
use. 

EXAMPLES    FOR    PRACTICE. 

1.  How  many  poles  in  a  township  of  land  ? 

2.  Reduce  3686400  P.  to  sq.  mi. 

3.  In  94  A.  7  sq.  ch.  12  P.    118  sq.  1.  how  many  square 
links  ? 

4.  Wimt  will  be  the  cost  of  a  farm  containing    4")r)0000 
square  links,  at  $<50  per  acre  ?  Ans.     $22750. 

Repeat  the  table  of  surveyors'  square  measure.     Give  the  scale. 


COMrOUXD   XUMBEKS. 


165 


III.     Cubic  Measure. 


^^=^z—zi 


y^ 


C' — 


^ 


-i^d. 


t9S.  A  Cube  is  a  polkl,  or  body, 
having  six  equal  square  sides,  or 
faces.  If  eacli  side  of  a  cube  be  1 
yard,  or  3  feet,  1  foot  in  thickness 
of  tliis  cube  Avill  contain  3X3X1 
v=.  9  cubic  feet, and  the  whole  cube  will 
contain  3  X  3  X  3  =r  27  cubic  feet. 
A  solid,  or  body,  may  have  the 
three  dimensions  all  alike  or  all  different.  A  body  4  ft.  long, 
3  ft.  wide,  and  2  ft.  tliick  contains  4  X  3  X  2  =:  24  cubic  or 
solid  feet.     Hence  we  see  that 

llic  cubic  or  solid  contents  of  a  body  are  found  by  midtipJy- 
iny  the  length,  breadth,  and  thickness  together. 

lf5S>.  Cubic  Measure,  also  called  Solid  Measure,  is  used 
in  estimating  the  contents  of  solids,  or  bodies  ;  as  timber,  wood, 
stone,  &c. 


TABLE. 


1728  cubic  inches  (cu.  in.)  make  1  cubic  foot,. . 

"       1  cubic  yard,. 
"       1  cord  foot, . . . 


27  cubic  feet 
16  cubic  feet 
8  cord  feet,  or 
128  cubic  feet 


!et,  or  ? 
;eet,     I 

24|-  cubic  feet 


..cu.  ft. 
.cu.  yd. 

.cd.*^ft. 

1  cord  of  wood,.  . .  .Cd. 

,  S  perch  of  stone  ?  -r>  i 
1  <  ^  >  Pch. 

;  or  masonry,      ^ 


Scale  —  ascending,  1728,  27.  Tae  other  numbers  are  not  in  a 
regular  scale,  but  are  merely  so  many  times  1  foot.  The  unit 
equivalents,  being  fractional,  are  consequently  omitted. 

NoTr,=;.     1.  A  cubic  yard  of  earth  is  called  a  load. 

2.  Railroad  and  transijortation  companies  estimate  li^lit  freight  hj 
the  space  it  occupies  in  cubic  feet,  and  hea^^  freight  by  weit^ht. 

3.  A  pile  of  wood  8  feet  lonp:,  4  feet  wide,  and  4  feet  high,  contain'5 
1  cord ;   and  a  cord  foot  is  1  foot  in  len2;th  of  such  a  pile. 

4.  A  perch  of  stone  or  of  masonry  is  16.^  feet  long,  1.^  feet  ■wide,  and 
1  foot  hiofh. 


Pefine  a  cube.  How  are  the  contents  of  a  cube  or  reetan^nlar 
solid  found?  For  what  i?i  cubic  measure  used?  Repeat  the  table. 
Give  the  scale.  How  is  railroad  freight  estimated  ?  "What  is  under- 
titood  by  a  cord  foot  ?     By  a  perch  of  stone  or  masonry  i 


166  REDUCTION. 

5.  Joiners,  bricklayers,  and  masons  make  no  allo-vvancc  for  -windows, 
doors,  &e.  liricklayers  and  masons,  in  estimating  their  -work  by 
cubic  measure,  make  no  allowance  for  the  corners  of  the  Avails  of 
houses,  Cellars,  &c.,  but  estimate  their  woi^k  by  the  girt,  that  i.-,  the 
entii-e  length  of  the  wall  on  the  out.-ide. 

6.  Engineers,  in  making  estimates  for  excavations  and  embankments, 
take  the  dimensions  with  a  line  or  measure  divided  into  fiet  and  deci- 
mals of  a  foot.  The  estimates  are  made  in  feet  and  decimals,  and  the 
results  are  reduced  to  cubic  yards. 

EXAMPLES    FOB    PRACTICE. 

,     1.    Inl25cu.ft.  840cu.in.ho\vmanycu.  in.  ?  ^hs.  216840. 

2.  Reduce  5224  cubic  feet  to  cords.  Ajis.   40||. 

3.  In  a  solid,  3  ft.  2  in.  long,  2  ft.  2  in.  -wide,  and  1  ft.  8  in. 
thick,  how  many  cubic  inches?  A?is.    19760. 

4.  How  many  small  cubes,  1  inch  on  each  edge,  can  be 
sawed  from  a  cube  6  feet  on  each  edge,  allowing  no  Avaste  for 
sawing?  Ans.    373248. 

5.  In  a  pile  of  wood  60  feet  long,  20  feet  wide,  and  15  feet 
liigh,  how  many  cords  ?  Ans.    140|, 

6.  IIow  many  cubic  feet  in  a  load  of  wood  10  feet  long,  3^ 
feet  Avide,  and  3 J-  i'eet  high  ?  Ans.    113  J  cu.  ft. 

7.  If  a  load  of  wood  be  12  feet  long  and  3  feet  wide,  hoAv 
high  must  it  be  to  make  a  cord?  Ans.    3|  ft.  high. 

8.  The  gray  limestone  of  Central  New  York  weighs  175 
pounds  a  cubic  foot.     What  is  the  weight  of  one  solid  yard  ? 

Alls.    2  T.  7  cwt.  25  lb. 
n.    A  cellar  wall,  32  ft.  by  24  ft.,  is  6  ft.  high  and  U  ft.  thick. 
IIow  much  did  it  cost  at  $1.25  a  perch?      Atis.    $50,909+ 

10.  IIow  much  did  it  cost  to  dig  the  same  cellar,  at  15 
cents  a  cubic  yard  ?  Aiis.    $2o.60. 

1 1 .  l\ry  sleeping  room  is  10  ft.  long,  9  ft.  wide,  and  8  ft.  high. 
Tfl  brcnthe  10  on.  ft. of  air  in  one  minute,  in  how  long  a  time  will 
I  breathe  as  much  air  as  the  room  contains  ?        Ajis.    72  min. 

12.  In  a  sciiool  room  30  ft.  long,  20  ft.  wide,  and  10  ft.  high, 
with  50  persons  breathing  each  10  cu.  ft.  of  air  in  one  minute, 
in  how  long  a  time  will  they  breathe  as  much  as  the  room 
contains?  Ans.    12  min. 

IIow  are  excavations  and  embankments  measured  ? 


COMPOUND   NUMBERS.  167 

MEASURES   OF   CAPACITY. 

I.     Liquid  Measure. 

SOO.    liquid  Measure,  also  called  Wine  Measure,  is  u-^ed 
in  measuring  liquids  ;  as  liquors,  molasses,  water,  &c. 

TABLE. 

4    gills  (gi.)  make  1  pint, pt. 

2    pints  "  1   q^i'iit, qt. 

4    quarts  "  1  gallo" g^l. 

3U  gallons  "  1   barrel, bbl. 

2    barrels,  or  63  gal.     "  1  hogshead,,  .hhd. 

XTNIT    EQUIVALENTS. 

pt.  .        pi. 

qt.  1=4 

,,1.  1  =       2  =r         8 

bW.         1    =       4  =       8  =       32 

],hd.        1  =  3U  =:  126  =  252  =  1008 

1  r=  2  =:  63    =r  252  zrr  504  =r  2016 

Scale  — ascending,  4,  2,  4,  31^,  2;  descending,  2,  311  4,  2,4. 
The  following  denominations  are  also  in  use : 


o 


36  gallons  make  1  barrel        of  beer. 
54       "        or  li  barrels  "     1  hogshead  "     " 

42       "  "     1  tierce. 

2  hogsheads,  or  120  gallons,        "     1  pipe  or  butt. 

2  pipes  or  4  hogsheads,  "     1  tun. 

Notes.  1.  The  denominations,  barrel  and  hogshead,  are  used  in  es- 
timating the  capacity  of  cisterns,  reservoirs,  vats,  &c. 

2.  The  tierce,  hogshead,  pipe,  butt,  and  tim  are  the  names  of  casks, 
and  do  not  express  any  fixed  or  definite  measures.  They  are  usually 
ganged,  and  have  their  capacities  in  gallons  marked  on  them. 

3.  Ale  or  beer  measure,  formerly  used  in  measuring  beer,  ale,  and 
milk,  is  almost  entirely  discarded. 

What  is  Uquid  measure  ?  Repeat  the  table.  Give  the  scale.  "WTiat 
other  denominations  are  sometimes  used  ?  How  are  the  capacities  of 
cisterns,  reservoirs,  &c.,  reckoned  ?     Of  large  casks  ? 


163 


REDUCTION. 


EXAMPLES    FOR    PRACTICE. 


1.   In  2  hhd.  1  bar.  30  gal.  2 
qt.  1  jjt.  3  gi.  how  many  gills  ? 

OPERATION. 

2  hhd.  1  bar.  30  gal.  2  qt. 
J_  [Ipt.  3gi. 

obbl. 
31J- 


185 

1871  gal. 
4 


752  qt. 
2 


1505  pt. 

4 


2.    In  G023  gi.  how  many 
hhds.  ? 

OPERATION. 

4 )  6023  gi. 

2  )  1505  pt.  +  3  gi. 

4  )752  qt.  +  1  pt. 

311    188  gal. 
2   J      2 


63   )376 


[gal. 


2)_5bbl.-f--V-gal.:=30^ 
2  hhd.  -f  1  bar. 

Jns.    2  hhd.  1  bar.  30i  gal. 

1  pt.  3  gi. 

But  ^  gal.  =r  2  qt.,  making 
the  A?is.  2  hhd.  1  bar.  30  gal. 

2  qt.  1  pt.  3  gi. 


6023  gi.,  A)is. 

3.  Reduce  3  hogsheads  to  gills. 

4.  Reduce  6048  gills  to  hogsheads. 

5.  In  13  hhd.  15  gal.  1  qt.  how  many  pints? 

6.  In  6674  pints  how  many  hogsheads  ? 

7.  What  will  be  the  cost  of  a  hogshead  of  wine,  at  6  cents 
a  gill?  Ans.    $120.96. 

8.  A  grocer  bought  1 0  barrels  of  cider,  at  $2  a  barrel ; 
after  converting  it  into  Ainegar,  he  retailed  it  all  at  5  cents  a 
quart ;  how  much  was  his  whole  gain  ?  Ans.    $43. 

9.  At  6  cents  a  pint,  how  much  molasses  can  be  bought  for 
$3.84?  Ans.   8  gal 

10.  How  many  demijohn?,  that  will  contain  2  gal.  2  qt.  1  pt. 
each,  can  be  filled  from  a  hogshead  of  wine  ?  Ans.    24. 

ir.     Dry  INIeasure. 

20  Bo  Dry  Measure  is  used  in  measuring  articles  not 
liquid,  as  grain,  fruit,  salt,  roots,  ashes,  &;c. 


"NMiat  is  dry  measure  ? 


COMPOUND   NUMBERS.  169 

TABLE. 

2  pints  (pt.)  make  1  quart, qt. 

8  quarts     "   1  peck, pk. 

4  pecks     "   1  bushel, .  bu.  or  bush. 

UNIT    EQUIVALENTS, 
qt.  pt. 

pk.  1=2 

bu.       1  =     8  =  16 
1  r=  4  =z  32  zr:  64 

Sc.VLE  —  ascending,  2,  8,  4  ;   descending,  4,  8,  2. 

Note.  In  England,  8  bu.  of  70  lbs.  each  are  called  a  quarter,  used  in 
mcasuiing  grain.    The  weight  of  the  English  quarter  is  1  of  a  long  ton. 

EXAMPLES    FOR    PRACTICE. 

1.  In  49  bu.  3  pk.  7  qt.  1  pt.  how  many  pints? 

2.  In  3199  pt.  how  many  bu.shels  ? 

3.  Ileduce  1  bu.  1  pk.  1  qt.  1  pt.  to  pints. 

4.  Reduce  83  pints  to  busheLs. 

5.  An  innkeeper  bought  a  load  of  50  bushels  of  oats  at  65 
cents  a  bushel,  and  retailed  them  at  25  cents  a  peck ;  how 
much  did  he  make  on  the  load  ?  Ans.   $17.50. 

STANDARD    OF    EXTENSION. 

S©?5.  The  U.  S.  standard  unit  of  measures  of  extension^ 
whether  linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or  36 
inches,  and  is  the  same  as  the  imperial  standard  yard  of 
Great  Britain.  It  is  determined  as  follows  :  The  rod  of  a 
pendulum  vibrating  seconds  of  mean  time,  in  the  latitude  of 
London,  in  a  vacuum,  at  the  level  of  the  sea,  is  divided  into 
391393  equal  parts,  and  360000  of  these  parts  are  36  inches, 
or  1  standard  yard.  Hence,  such  a  pendulum  rod  is  39.1393 
inches  long,  and  the  standard  yard  is  ff^jf^^  of  the  length  of 
the  pendulum  rod. 

203.  The  U.  S.  standard  unit  of  liquid  measure  is  the  old 
English  wine  gallon,  of  231  cubic  inches,  which  is  equal  to 
8.83888  pounds  avoirdupois  of  distilled  water  at  its  maximum 
density,  that  is,  at  the  temperature  of  39.83°  Fahrenheit,  the 
barometer  at  30  inches. 

Repeat  the  table.     "What  is  a  quarter  ?     What  is  the  U.  S.  standard 
unit  of  measurement  of  extension  ?     How  is  it  determined  f     What  is 
the  U.  S.  standard  unit  of  liquid  measure  ? 
R.P.  8 


170  REDUCTION. 

^O^t.  TJie  U.  S.  standard  unit  of  dry  measured  the  Brit- 
ish ^ViIlche.ster  bushel,  wliich  is  184-  inelies  in  diameter  and  8 
inches  deep,  and  contains  2150.42  cubic  inches,  equal  to 
77.6274  pounds  avoirdupois  of  distilled  water,  at  its  maximum 
density.     A  gallon,  dry  measure,  contains  2G8.8  cubic  inches. 

Note.  1.  The  wine  and  drj' measures  of  the  same  denomination 
are  of  different  capacities.  The  exact  and  the  relative  size  of  each  may 
be  readily  seen  by  the  following 

3|0»5.    COMPARATIVE    TABLE    OP  MEASURES    OF  CAPACITY. 

Cn.  in.  in         Cu.  in.  in         Cu.  in.  in  Cu.  in.  in 

cue  i;all(jn.      oue  quart.       one  pint.  one  gill. 

Wine  measm-e,  231  57|  28|  TgV 

Dry  measiu-e,  (\  pk.,)  26S|  Q>~i\  33|  8f  " 

2.  The  beer  gallon  of  282  inches  is  retained  in  use  only  by  custom 
A  bushel  is  commonly  estimated  at  21504  cubic  inches. 

EXAMPLES    FOR    PHACTICE. 

1.  A  fruit  dealer  bought  a  bushel  of  strawberries,  dry 
measure,  and  sold  them  by  wine  measure  ;  how  many  quarts 
did  he  gain  ?  Ans.    o^f  quarts. 

2.  A  grocer  bought  40  quarts  of  milk  by  beer  measure,  and 
sold  it  by  wine  measure ;  how  many  quarts  did  he  gain  ? 

Aris.    8f  f  quarts. 

3.  A  busheh  or  32  quarts,  dry  measure,  contains  how  many 
more  cubic  inches  than  32  quarts  wine  measure  ? 

Ans.    302f  cu.  in. 
Tlme. 
S06.    Time  is  used  in  measuring  periods  of  duration,  as 
years,  days,  minutes,  &c. 

T.\r.i,E. 

60  seconds  (sec.)   make  1  minute, min. 

60  minutes  "       1  hour h. 

21  hours  "      1   day, da. 

7  days  "      1  week, wk. 

365  days  "      1  common  year,. .  .yr. 

366  days  "      1  leap  year, yr. 

12  calendar  months   "      1  year, yr. 

100  years  "      1  century C. 

"What  is  the  U,  S.  standard  unit  of  dry  measure  ?  How  is  it  ob- 
tained ?  "What  is  the  relative  sizo  of  the  wine  and  the  dry  p:nllon  ? 
Wliat  is  the  .size  of  a  beer  gallon  ?     What  is  time  ?     Ilcpcat  the  tabic. 


COMrOUND   NUMBERS. 


171 


UNIT    EQUIVALENTS. 


mm. 

sec. 

1,.                       1    = 

60 

d..                1   =            60  := 

3600 

wk.               1  =       24  =       1440  = 

SG'IOO 

1  =         7  =     168  =:     10080  — 

604800 

y 

mo.        ^  365  z=  87GO  z=  525600  = 

31536000 

1 

—  12  —  )  366  —  8784  —  527040  — 

31622400 

Scale  —  ascending,  60,  60,  24,  7  ;  descending,  7,  24,  60,  60. 

The  calendar  year  Is  divided  as  follows :  — 

Names. 

January, 

February, 
'  March, 

April, 
.  May, 
'  June, 
!  July, 
'  August, 
r  September, 
?  October, 
I  November, 

December, 


No.  of  nio. 

Sp;iPon, 

1 

Winter, 

2 

it 

3 

Spring, 

4 

(. 

5 

a 

6 

Summer, 

7 

(t 

8 

« 

9 

Autumn, 

10 

« 

11 

It 

12 

Winter, 

AbbreTiations. 

Jan. 
Feb. 
Mar. 
Apr. 

Jun. 

Sept. 
Oct. 
Nov. 
Dec. 


No.  of  days. 

31 

28  or  29 

31 

30 

31 

30 

31 

31 

30 

31 

30 

31 


365  or  366 


Notes.  1.  The  exact  length  of  a  solar  year  is  365  da.  5  h.  48  min.  46 
Bee. ;  hut  for  convenience  it  is  reckoned  II  min.  14  sec.  more  than  this, 
or  3G5  da.  6  h.  =  30.5^  da.  This  \  day  in  4  years  makes  one  day, 
which,  every  fourth,  bissextile,  or  leap  year,  is  added  to  the  shortest 
month,  givins;  it  29  days.  The  leap  y^ars  are  exactly  divisible  by  4, 
as  1856,  1860,  1864.  The  number  of  days  in  each  calendar  month 
may  be  caisily  remembered  by  committing  the  followmg  lines  :  — 

"  Thirty  flays  hath  Si'iitfinher, 
April,  Junp.  ai)il  November; 
AJl  the  rt'st  liavp  tliirty-one, 
Pave  February,  which  alono 
Hath  twenty-eiglit ;  and  one  day  more 
AVe  add  to  it  one  year  in  four." 

2.  In  most  business  transactions  30  days  are  called  1  month. 
EXAjrPLES    FOR    PRACTICE. 

1.  Reduce  365  da.  5  h.  48  min.  46  sec.  to  seconds. 

2.  Reduce  315.36926  seconds  to  days. 

Give  the  scale.  "What  is  the  len<;tli  of  each  of  the  calendar  months  ? 
What  is  the  exact  length  of  a  solar  year  ?  Explain  the  use  of  bissextile 
•r  leap  year.     What  is  the  length  of  a  month  in  busiiiess  transactions? 


172  KEDUCTION. 

3.  In  5  wk.  1  da.  1  li.  1  min.  1  sec.  how  many  seconds  ? 

4.  In  31140G1  seconds  how  many  weeks? 

5.  How  many  times  does  a  clock  pendulum,  3  ft.  3  in.  long, 
beating  seconds,  vibrate  in  one  day?  Ans.     86400. 

6.  If  a  man  take  1  step  a  yard  long  in  a  second,  in  how 
long  a  time  will  he  walk  10  miles  ?  Ans.   4  h.  53  min.  20  sec. 

7.  In  a  lunar  month  of  29  da.  12  h.  44  min.  3  sec.  how 
many  seconds?  Ans.   2551443. 

8.  How  much  time  will  a  person  gain  in  40  years,  by  rising 
45  minutes  earlier  eveiy  day  ?      Ans.  456  da.  13  h.  30  min. 

Circular  Measure. 

S®7.  Circular  Measure,  or  Circular  Motion,  is  used  prin- 
cipally in  surveying,  navigation,  astronomy,  and  geography, 
for  reckoning  latitude  and  longitude,  determining  locations  of 
places  and  vessels,  and  computing  difference  of  time. 

Every  circle,  great  or  small,  is  divisible  into  the  same  num- 
ber of  equal  parts,  as  quarters,  called  quadrants,  twelfths, 
called  signs,  360lhs,  called  degrees,  &c.  'Consequently  the 
parts  of  different  circles,  although  having  the  same  names,  are 
of  different  lenjrths. 


c 


TABLE. 


60  seconds  (")      make  1  minute, .  . .  '. 

60  minutes  "  1  decree, . . .  ^. 

30  doi^rees  "  1  Kia:n S. 

12  signs,  or  3G0°,    "  1  circle, C. 

UNIT    EQUIVALENTS. 

/  /' 

1  =  60 

P.  1  r=         60  r=       .  3600 

r.         1  m    30  —    i.soo  :=    insnoo 

1  =  12  =  360  —  21600  =:  1296000 
Scale  —  ascending,  60,  60,  30,  12;  descending,  12,  30,  60,  60. 

Notes.     1.    Minutes  of  the  earth's  circumference  are  called  geo- 
graphic or  nautical  mile?:. 

2.  The  denomination,  sir/us,  is  confined  exclusively  to  Astronomy. 

Define  circular  measure.      How  are  circles  rlividod  ?      llcpcat  the 
table.      Give  the   scale.      "What   is  a  geographic   mile  ?      "What  i*  S  i| 
sign  ? 


COMPOUND   NUMBERS.  •  ITS 

3.  Degrees  are  not  strictly  divisions  of  a  circle,  but  of  tlie  space 
about  a  point  in  any  pbne. 

4.  90°  make  a  quadrant,  or  right  angle,  and  G0°  a  sextant,  or  -I  of  a 
cii'cle. 

EXAMPLES    FOll    PRACTICE. 

1.  Reduce  10  S.  10°  10'  10"  to  seconds. 

2.  Reduce  lUGGlO''   to  signs. 

3.  How  many  degrees  in  11 -100  geograpliic  or  nautical 
r.:iles?  Ans.    190°. 

4.  If  1  degree  of  the  earth's  circumference  is  69^  statute 
miles,  how  many  statute  miles  in  11-100  geograpliic  miles,  or 
190  degrees?  A7is.  13148. 

5.  How  many  minutes,  or  nautical  miles,  in  the  circum- 
ference of  the  earth?  Ans.  21600'  or  mi. 

G.  A  ship  during  4  days'  storm  at  sea  changed  her  longitude 
397  geographical  miles;  how  many  degrees  and  minutes  did 
she  cluuige  ?  Ans.   6°  37'. 

2®8.      In  Counting. 

12  units  or  things. . .  .make. ...  1  dozen. 
12  dozen  "  1  gross. 

12  gross  "  1  great  gross. 

20  units  "  1  score. 

S®!>.     Paper. 

24  sheets make 1  quire. 

20  quires  "  1  ream. 

2  reams  "  1  bundle. 

5  bundles  "  1  bale. 

SiO.      Books. 

The  terms  folio,  quarto,  octavo,    duodecimo,  &c.,  indicate 

the  number  of  leaves  into  Avhich  a  sheet  of  paper  is  folded. 

A  sheet  folded  in    2  leaves  is  called  a  folio. 

A  sheet  folded  in    4  leaves  "  a  quarto,  or  4to. 

A  sheet  folded  in    S  leaves  "  an  octavo,  or  8vo. 

A  sheet  folded  in  12  leaves  "  a  12mo. 

A  sheet  folded  in  16  leaves  "  a  IHmo. 

A  sheet  folded  in  IS  leaves  "  an  ISmo. 

A  sheet  folded  in  24  leaves  "  a  24010. 

A  sheet  folded  in  32  leaves  "  a  32mo. 

A\Tiat  is  a  deo;ree  ?  Repeat  the  table  for  counting.  For  reckoning 
paper.    For  indicating  the  size  of  books. 


f,  4  REDUCTION. 

EXAMPLES    FOR    PRACTICE. 

1.  If  in  Birmingliam,  England,  150  million  Gillott  pens  are 
manufactured  annually,  liow  many  great  gross  will  tliey  make  ? 

Ans.    SG8U5  great  gross  6  gross  8  dozen. 

2.  In  100000  sheets  of  paper,  how  many  bales? 

Aiis.    20  bales  4  bundles  6  quires  16  sheets. 

3.  What  is  the  age  of  a  man  4  score  and  10  years  old? 

4.  How  many  printed  pages,  2  pages  to  each  leaf,  will 
there  be  in  an  octavo  book,  having  8  fully  printed  sheets  ? 

A71S.    128  pages. 

5.  How  large  a  book  will  ten  32mo.  sheets  make,  if  every 
page  be  printed?  Ans.    G40  pages. 

PROMISCUOUS    EXAMPLES    IX    REDUCTION. 

1.  How  many  suits  of  clothes,  each  containing  6  yd.  3f  qr., 
can  be  cut  from  333  yards  of  cloth  ?  Ans.    48. 

2.  A  man  bought  a  gold  chain,  weighing  1  oz.  15  pwt.,  at 
seven  dimes  a  pennyweight;  what  did  it  cost?      Ans.  $24. cO. 

3.  A  physician,  having  2  tb  3§  5  5  19  10  gr.  of  medicine, 
dealf  it  out  in  prescriptions  averaging  15  grains  each;  how 
many  prescriptions  did  it  make  ?  Ans.    886. 

4.  A  man  bought  1  T.  11  cwt.  12  lbs.  of  hay,  at  IJ-  cents 
a  pound  ;  wliat  did  it  cost.''  Ans.    Ci^38.90. 

5.  What  will  be  the  cost  of  a  load  of  oats  weighing  1456 
pounds,  at  37^  cents  per  bushel  ?  AicS.    $17.0625. 

6.  If  one  bushel  of  wheat  will  make  45  pounds  of  flour,  how 
many  barrels  will  lOOO  bushels  make  ?     Ans.  229  bbl.  116  lb. 

7.  A  load  of  wheat  weighing  2430  pounds  is  worth  how 
much,  at  $1.20  a  bushel?  Ans.    $48.60. 

8.  Paid  $12.50  for  a  barrel  of  beef;  how  much  was  that 
per  pound?  Ans.    6^  cents. 

0.  If  a  silver  dollar  measure  one  inch  in  diameter,  how 
many  <lullars,  laid  side  by  side  on  the  e(iuator,  would  reach 
round  the  earth?  Ans.    l.")73862400. 

10.    In  10  mi.  7  cli.  4  rd,  201.,  how  many  links? 

Ans.    80820  Hnks. 


DENOMINATE   FRACTIONS.  175 

il.  What  is  the  value  of  a  city  lot,  25  feet  wide  and  100 
feet  long,  if  every  square  inch  is  worth  one  cent?   Aiis.  S;)60(). 

12.  How  many  cords  of  wood  can  be  ])ik'd  in  a  shed  50  I't. 
long,  25  ft.  wide,  and  10  ft.  high  ?      Atis.  97  Cd.  5  cd.  ft.  4  cu.  ft. 

13.  A  cistern  10  feet  square  and  10  feet  deep,  will  hold 
how  many  liogsheads  of  water?     Ans.  118  hhd.  46^^  gal. 

14.  A  bin  8  feet  long,  5  feet  wide,  and  4J  feet  high,  will 
hold  how  many  bushels  of  grain  ?  Ans.    144y^j  bu. 

15.  How  many  seconds  less  in  every  Autumn  than  in 
either  Spring  or  Summer?  Ans.   86400  sec. 

16.  If  a  person  could  travel  at  the  rate  of  a  second  of  dis- 
tance in  a  second  of  timCj  how  much  time  would  he  require  to 
travel  round  the  earth?  Ans.    15  days. 

17.  How  many  yards  of  carpeting,  1  yd.  wide,  will  be  re- 
quired to  carpet  a  room  20  ft.  long  and  18  ft.  wide  ?    Ans.  40. 

18.  A  printer  calls  for  4  reams  10  quires  and  10  sheets  of 
paper  to  print  a  book ;  how  many  sheets  does  he  call  for  ? 

Ans.    2170. 

19.  How  many  times  will  a  wheel,  16  ft.  6  in.  in  circumfer- 
ence, turn  round  in  running  42  miles?  Ans.    13440. 

20.  How  many  days,  working  10  hours  a  day,  will  it  re- 
quire for  a  person  to  count  $10000,  at  the  rate  of  one  cent 
each  second?  Ans.    27  da.  7h.  46  min.  40  sec. 

21.  A  town,  6  miles  long  and  4^  miles  wide,  is  equal  to 
how  many  farms  of  80  acres  each  ?  Ans.    216. 

22.  At  $21.75  per  rod,  what  will  be  the  cost  of  grading 
10  mi.  176  rds.  of  road  ?  Ans.  $73428. 


EEDUCTION   OF   DENOMINATE   FRACTIONS. 
CASE    I. 

Sll,     To   reduce    a   denominate    fraction   from  a 
greater  to  a  less  unit. 

1.    Reduce  ^'g-  of  a  bushel  to  the  fraction  of  a  pint. 

Case  I  is  what  ? 


176  '  EEDUCTION. 


OPERATION. 

1    V4V8VS  —  4  A„e          Analysis,     lo  reduce  bushels 

Q  to  pints,  we  must  multijjiy  by  4, 

r           '  8,    and   2,   the    numbers   in    the 

00     1  scale.     And  since  the  given  nura- 

4  ber  is  a  fraction  of  a  bushel,  -vve 

ri  indicate  the  process  as  in  multi- 

ci  plication   of  fractions,   and    after 

canceling,  obtain  -f,  the  Answer. 

4  =:  I  pt.,  Ans.     Hence, 


Rule.  Multipli/  the  fraction  of  the  higher  denomination  by 
the  numhers  in  the  scale  successively,  between  the  given  and  tlte 
required  denominations. 

Note.     Cancellation  may  be  applied  wherever  practicable. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  y^Va  of  a  £  to  the  fraction  of  a  penny. 

Ans.   -rp^  d. 

3.  Reduce  xi itro-  of  a  week  to  the  fraction  of  a  minute. 

Ans.    -^^  min. 

4.  "What  part  of  a  gill  is  j^fW  ^^  ^  hosghead  ?   Ans.   \  gi. 

5.  What  fraction  of  a  grain  is  ^^^^  of  an  ounce  ?    Ans.  ^  gr. 

6.  Reduce  x^oioiya-  ^^  ^  ^i^^  to  ^^^^  fraction  of  an  inch. 

Ans.    Jt%^-  in. 

7.  Reduce  f  of  ^  of  2  pounds  to  the  fraction  of  an  ounce 
Troy.  Ans.  f  oz. 

8.  Reduce  ^^-^  of  a  hogshead  to  the  fraction  of  a  pint. 

Ans.    11  pt. 

9.  Reduce  xiTty  ^^  ^^^  acre  to  the  fraction  of  a  rod. 

Ans.  ^  rd. 

CASE   II. 

213.  To  rccIucG  a  denominate  fraction  from  a  less 
to  a  f^rcater  unit. 

1.    Reduce  ^  of  a  pint  to  the  fraction  of  a  bushel. 
Give  explanation.     Rule.     Case  It  is  what  ? 


DENOMINATE    FRACTIONS.  177 


OPERATION. 


^1111 

X  — X— X  — =— ,  Ans. 


Analysis.       To      reduce 
pints    to   bushels,    Ave    must 


5         2         S         4  ~80'  '        ^livide   by  2,  8,   and    4,   the 

numbers  of  the  scale.     And 
4  since   the    given    number  of 

pints  is  a  fraction,  we  indi- 
cate the  process,  as  in  divis- 
ion  of  fractions,  and  cancel- 


Or,      5 
o 

8 
4 


80 


1  =  ^L  bu.,  Ans. 


ing,  obtain  Jg,  the  Answer. 


Rule.  Divide  t/te  fraction  of  the  lower  denomination  hy  the 
numbers  in  the  scale,  successively,  between  the  given  and  the 
required  denomination. 

Note.     The  operation  Avill  frequently  be  shortened  by  cancellation. 
EXAMPLES    FOR    PKACTICE. 

2.  What  part  of  a  rod  is  ^  of  a  foot  ?  Ai^s.   -^^^  rd. 

3.  What  part  of  a  pound  is  -^  of  a  dram?       Ans.    -^^^^  lb. 

4.  Reduce  4  of  a  cent  to  the  fraction  of  an  eagle. 

Ans.    j()VtT  E. 

5.  A  hand  is  ^  of  a  foot ;  what  fraction  is  that  of  a  mile  ? 

6.  Reduce  f  of  2  pwt.  to  the  fraction  of  a  pound.  Ans.  ^^jj  lb. 

7.  How  much  less  is  f  of  a  pint  than  2-  of  a  hogshead  ? 

Ans.    1 19-  hhd. 

8.  In  f  of  an  inch  what  fraction  of  a  mile  ?  Ans.  y^-jVoff  "^i- 

9.  ^  of  an  ounce  Tioy  is  f  of  what  fraction  of  2  pounds  ? 
10.    Tj  of  an  ounce  is  ^  of  what  fraction  of  2  pounds  Troj? 

CASE    III. 

2!13.  To  reduce  a  denominate  fraction  to  integers 
of  lower  denominations. 

1.    What  is  the  value  off  of  a  hogshead  of  wine? 

Give  explanation.     Rule.     Case  III  is  what  ? 
8« 


173 


REDUCTION. 


OPERATION. 

I-  Lhcl.  X  63  =  3.1 5  gal.  zrr  39f  gal. 
i  gal.  Xi  =  J/-  qt.  :=  1 1-  qt. ;   |  qt.  X  '2  =  f  pt.  =  1  pt. 

A?is.    39  gal.  1  qt.  1  pint. 

Analysis.  |  hhd.  :=  f  of  63  gal.,  or  39|  gal. ;  and  |  gal.  =  |  of 
4  qt.,  or  1|  qt.  ;   and  |  qt.  ^  |  of  2  pt.,  or  1  pt.     Hence, 

Rule.  I.  3IuUipli/  the  fraction  by  that  numher  in  the  scale 
lohich  loill  reduce  it  to  the  next  lower  denomination,  and  if  the 
result  be  an  improper  fraction,  reduce  it  to  a  whole  or  mixed 
number. 

II.  Proceed  with  the  fractional  part,  if  any,  as  before, 
until  reduced  to  the  denominations  required. 

III.  The  units  of  the  several  denominatio7is,  arranged  in 
their  order,  will  be  the  required  result. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  f-  of  a  month  to  lower  denominations. 

Ans.    17  da.  3  h.  25  min.  421  sec. 

3.  TThat  is  the  value  of  ^  of  a  £  ?         Ans.   8  s.  G  d.  ^  far. 

4.  AVhat  is  the  value  off  of  a  bushel  ? 

5.  Reduce  f  of  15  cwt.  to  its  equivalent  value. 

Ans.    12  cwt.  85  lbs.  11  oz.  ^  dr. 

6.  Reduce  f  of  |  of  a  pound  avoirdupois  to  integers. 

Ans.    4oz.  n#adr. 

7.  What  is  the  value  of  f  of  an  acre  ?       Ans.    3  R.  13^  P. 

8.  Reduce  4f  of  a  day  to  its  value  in  integers. 

Ans.    16  li.  3G  min.  So-j^-j-  sec. 

9.  What  is  the  value  of  |  of  a  pound  Troy  ? 

10.  What  is  the  value  of  |^  of  51  tons  ?  Ans.  4  T.  5  cwt.  55f  lb. 

1 1.  What  is  the  value  of  f  of  3§  acres  ?   Ans.  1  A.  1  R.  20  P. 

CASE    IV. 

31'!.    To  rciluco  a  compound  number  to  a  fraction 
of  a  liiglier  denomination. 

1.    What  part  of  a  week  is  5  da.  14  h.  24  min.  ? 


Give  explanation.     Rule.     Case  IV  is  what  ? 


DENOMINATE   FRACTIONS.  179 

oPEKATiox.  Analysis.    To  find 

5  da.  14  Ii.  21  mill.  :=  80  Gl  min.  what  part  one  compound 

1  Avk.  =  10080  min.  number  is    of  another, 

8  0  6  1   —  4  ,v!-      A  „<!  ^^'^y  "^^^^^  ^^  reduced  to 

•'^^'*^         "  the  same  denommatiou. 

In  o  da.  14  h.  24  min.  there  are  8064  minutes,  and  in  1  week  there 
are  10080  minutes.  Since  1  minute  is  iq^q^  of  a  week,  8004  min- 
utes is  /qVb'*o  =  -5  of  a  Aveek.     Hence, 

Rule.  Reduce  the  given  numher  to  its  lowest  denomination 
for  the  numerator,  and  a  unit  of  the  required  denomination 
to  the  same  denomination  for  the  denominator  of  the  required 
fraction. 

Note.  If  the  given  nmnber  contain  a  fraction,  the  denominator  of 
this  fraction  must  be  regarded  as  the  lowest  denomination. 

EXAMPLES    FOR    PKACTICE. 

2.  What  part  of  a  mi.  is  6  fur.  2G  rd.  3  yd.  2  ft.  ?   Ans.  f  mi. 

3.  "What  fraction  of  a£  is  13  s.  7  d.  3  far.? 

4.  Reduce  10  oz.  lOpwt.  10  gr.  to  the  fraction  of  a  pound 
Tioy.  Ans.   faf  lb. 

5.  Reduce  2  cd.  ft.  8  cii.  ft.  to  the  fraction  of  a  cord. 

Ans.    j5g.  Cd. 

0.  Reduce  1  bbl.  1  gab  1  qt.  1  pt.  1  gi.  to  the  fraction  of  a 
hogshead.  Ans.  J4|-  blid. 

7.  What  part  of  2  rods  is  4  yards  IJ-  feet?  Ans.   ■^^. 

8.  Reduce  1-|  peeks  to  the  fraction  of  a  busheb    Ans.  §  bu. 

9.  What  part  of  9  feet  square  are  9  square  feet  ? 

10.    From  a  piece  of  cloth  containing  8  yd.  3  qr.  a  tailor  cut 
2  }d.  2  qr. ;  Avbat  part  of  the  whole  piece  did  he  take  ?    Aa^.  f. 

CASE   V. 

Sl»>.  To  reduce  a  denominate  decimal  to  integers 
of  lower  denominations, 

1.  Reduce  .78125  of  a  pound  Troy  to  integers  of  loiver  de- 
nominations. 

Give  explanation.     Rule.     Case  V  is  what  ? 


180 


REDUCTION. 


Analysis.  We  first  multiply 
by  12  to  reduce  the  given  number 
from  pounds  to  ounces,  and  the 
result  is  9  ounces  and  the  decimal 
.375  of  an  oz.  We  then  multiply 
this  decimal  by  20  to  reduce  it  to 
pennyweights,  and  get  7  put.  and 
.5  of  a  pwt.  This  last  decimal  Me 
multiply  by  24,  to  reduce  it  to 
grains,  and  the  result  is  12  gr. 
Hence  the  answer  is  9  oz.  7  pwt. 
12  gr. 

Rule.     I.  Multiply  the  given  decimal  by  that  mimber  in  the 
scale  which  loill  reduce  it  to  the  next  lower  denomination^  and' 
point  off  as  in  multiplication  of  decimals. 

11.  Proceed  with  the  decimal  part  of  the  product  in  the  some 
manner  until  reduced  to  the  required  denominations.  The  in' 
tegers  at  the  left  will  he  the  answer  required. 


OPERATION. 

.78125  lb. 
12 

9.37500  oz. 
20 

7.50000  pwt. 
24 

12.0000  gr. 

9  oz.  7  pwt.  12  gr.,  Ans. 


2.  What  is 

3.  What  is 

4.  Reduce 
inations. 

5.  Reduce 

6.  What  is 

7.  What  is 

8.  What  is 

9.  What  is 
10.  W^hatis 


EXAMPLES    FOR    PRA.CTICE. 

the  value  of  .217°?  Ans.    13'  1.2''. 

the  value  of  .G59  of  a  week  ? 

Ans.    4  da.  14  h.  42  min.  43.2  sec. 
.578125  of  a  bushel  to  integers  of  lower  denom- 

Ans.    2  pk.  2  qt.  1  pt- 
.125  bbl.  to  integers  of  lower  denominations. 

Ans.    3  gal.  3  qt.  1  pt.  2  gi. 
the  value  of  .628125  £? 
the  value  of  .22  of  a  hogshead  of  molasses  ?. 

Ans.    13  gal.  3  qts.  3.52  gi. 

the  value  of  .G7  of  a  league  ? 

Ans.    2  mi.  3  rd.  1  yd.  3-^  in. 
the  value  of  .42857  of  a  month  ? 

Ans.    1 2  da.  20  h.  34  min.  IH^  sec. 
the  value  of  .78875  of  a  long  ton  ? 

Ans.    15  cwt.  3  qr.  2  lb.  12.8  oz. 


Give  explanation.    Rule. 


DENOMINATE   FRACTIONS.  181 

11.  Wliat  is  the  value  of  5.88125  acres  ?  Arts.  5  A.  3  R.  21  P. 

12.  Reduce  .0055  T.  to  pounds.  Ans.    11  lb. 

13.  Reduce   .034375   of  a  bundle  of  paper  to  its  value  in 
lower  denominations.  Ans.    1  quire  9  sheets. 

CASE    VI. 

SI6.  T6  reduce  a  compound  number  to  a  decimal 
of  a  higher  denomination. 

1.  Reduce  3  pk.  2  qt.  to  the  decimal  of  a  bushel. 

OPERATION,  Analysis.     Since  8  quarts  make 

8    2.00      qt.  1  peck,  and  4  pecks  1  bushel,  there 

T~7~  will  be  \  as  many  pecks  as  quarts 

J_lll!_ ^^"  (1§3),    and  ^  as  many  bushels  as 

.8125  bu.,  Ans.     pecks. 

Or  we  may  reduce  3  pk.  2  qt.  to 

Or,  3  pk.  2  qt.  —  2b  qt.         ^^^  fraction  of  a  bushel  (as  in  2 1 1 ), 

1  bu.         =  3-  qt.         and  we  have  |4  o^  ^  bushel,  M'hicli, 

1^  =  .8125  bu.,  Ans.      reduced  to  a  decimal,  equals  .8125. 

Hence  the 

Rule.  Divide  the  lowest  denomination  given  by  that  num- 
ber in  the  scale  which  icill  reduce  it  to  the  next  higher,  and  ttn- 
nex  the  quotient  as  a  decimal  to  that  higher.  Proceed  in  the 
same  manner  until  the  whole  is  reduced  to  the  denomination 
required.     Or,  ' 

Reduce  the  given  number  to  a  fraction  of  the  required  dc' 
nomination,  and  reduce  this  fraction  to  a  decimal. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  3  qt.  1  pt.  1  gi,  to  the  decimal  of  a  gallon. 

Ans.   .90G25ga1. 

3.  Reduce  10  oz.  13  pwt.  9  gr.  to  the  decimal  of  a  pound 
Troy.  Ans.  .88906251b. 

4.  Reduce  1.2  pints  to  the  decimal  of  a  hogshead. 

Ans.  .00238 +  hhd. 

5.  What  part  of  a  bushel  is  3  pk.  1.12  qt.  ?      Ans.  .785  bu. 

Case  VI  is  what  ?     Give  explanations.     Rule. 


IS2 


ADDITION. 


6.  What  part  of  an  acre  is  3  R.  12.56  P.  ? 

7.  Reduce  17  jd.  1  ft.  6  in.  to  the  decimal  of  a  mile. 

Ans.  .00994318 -f  mi. 

8.  Reduce  .32  of  a  pint  to  the  decimal  of  a  bushel. 

Ans.  .005  bu. 

9.  Reduce  A}  feet  to  the  decimal  of  a  fathom. 

Ans.  \8125  fothom. 

10.  Reduce  150  sheets  of  paper  to  the  decimal  of  a  ream. 

Ans.  .3125  Rm. 

11.  Reduce  47.04  lb.  of  flour  to  the  decimal  of  a  barrel. 

12.  Reduce  .33  of  a  foot  to  the  decimal  of  a  mile. 

13.  Reduce  5  h.  3G  min.  57y^jy  sec.  to  the  decimal  of  a  day. 

ADDITION. 

SST.  1.  A  miner  sold  at  one  time  10  lb.  4  oz.  IG  pwt.  8  gr. 
of  gold  ;  at  another  time,  2  lb.  9  oz.  3  pwt. ;  at  another,  11  oz. 
20  gr.  ;  and  at  another,  25  lb.  16  pwt.  23  gr.  ;  how  much  did 
he  sell  in  all  ? 

AtsULYSIS.  Arranging  the  num- 
bers in  columns,  placing  units  of  the 
sayne  denomination  under  each  otli- 
er,  wc  first  add  the  vniits  in  the 
right  hand  column,  or  lowest  de- 
nomination, and  find  the  amount  to 
be  51  grains,  which  is  equal  to  2 
pwt.  3  gr.  AYe  write  the  3  gr.  under 
the  column  of  grains,  and  add  tlie  2 
pwt.  to  the  column  of  pwt.  Wc  find  tlie  amount  of  the  second  col- 
unni  to  be  37  pwt.,  which  is  equal  to  1  oz.  17  ]nvt.  "Writing  the  17 
])wt.  under  the  column  of  pwt.,  we  add  the  1  oz.  to  the  next  column. 
Adding  this  column  in  the  same  manner  as  the  preceding  ones,  wc 
find  the  amount  to  be  25  oz.,  equal  to  2  lb.  1  oz.  Placing  the  1  oz. 
niiilcr  the  column  of  oz..  we  add  the  2  lb.  to  the  column  of  lb. 
Addintr  the  last  column,  we  find  the  amount  to  be  3911).  Hence 
the  following 


OPEUATION. 

IV.. 

oz.     pwt. 

gr. 

10 

4     16 

8 

2 

9       3 

0 

0 

11       0 

20 

25 

0     16 

23 

39 

1     17 

3 

"What  is  addition  of  compoimd  nvmibers  ?      Give  explanation. 


COMrOUND   NUMBERS.  183 

TtuLE.  I.  Write  (he  nuvibers  so  tJiat  tJiosc  of  the  same  unit 
VGiKe  will  stand  in  the  same  column. 

II.  Beginning  at  the  right  iiaiul,  add  each  denomination  as 
in  simple  numbers,  carrying  to  each  succeeding  denomination 
one  for  as  many  units  as  it  takes  of  the  denomination  added,  to 
make  one  of  t'ue  next  higher  denomination. 


EXAMPLES  FOR 

PRACTICE 

:. 

(-'•) 

( 

:-) 

£.   s.    d. 

lb. 

§• 

5-  9- 

gr- 

43  13   8 

12 

8 

7  2 

15 

51   6   4 

10 

4  1 

10 

67  11   3 

15 

00 

2  1 

19 

76  18  10 

11 

6  0 

12 

244  10   1 

13 

4 

4  2 

00 

(4.) 

(5.) 

T.  CAVt.  lb.   oz. 

cir. 

bu. 

pk.  qt. 

pt. 

4   7  IS   4 

10 

1 

3  7 

1 

15  98  15 

5 

3 

2  2 

0 

3   9  10   G 

15 

1  G 

1 

1   0  15   0 

4 

17 
45 

0  5 
2  4 

1 

9  12  42  11 

2 

0 

G.  What  is  tlie  sum  of  4  mi.  3  fur.  30  rd.  2  jd.  1  ft.  10  in,, 
5  mi.  Gfur.  18  rd.  1yd.  2  ft.  Gin.,  10  mi.  4  fur.  25  rd.  2  yd. 
2  ft.  11  in.,  and  G  fur.  28  rd.  4  yd.  2  ft.  1  in.  ?  ^ 

7.  Find  the  sum  of  197  sq.  yd.  4  sq.ft.  104J-  sq.  in.,  122 
sq.  yd.  2  sq.  ft.  27^  sq.  in.,  5  sq.  yd.  8  sq.  ft.  2f  sq.  in.,  and  237 
sq.  yd.  7  sq.ft.  128|sq.  in.? 

Ans.    563  sq.  yd.  4  sq.  ft.  118.825  sq.  in. 

Note.  When  common  fractions  occur,  they  shonlrT  be  reduced  to  a 
common  denominator,  to  decimals,  or  to  integers  of  a  lower  denomi- 
nation, and  added  according  to  the  usual  method. 

Give  the  Rule. 


18i 


1 

ADDITION. 

(S.) 

A. 

E. 

P. 

sq.  yd.  i 

;q.  ft.  s 

q.  in. 

26 

3 

28 

25 

8   1 

25 

19 

2 

38 

30 

7   1 

50 

456 

2 

20 

16 

6 

98 

503 

1 

8 

i2(i: 

)  5 

85 

W  = 

72 

503 

1 

8 

13 

1 

13 

(0.) 

(10.) 

mi. 

fur. 

rd. 

yd. 

ft. 

in. 

hhd 

.   gal. 

qt. 

pt. 

1 

7 

30 

4 

2 

11 

27 

65 

3 

2 

3 

4 

00 

2 

1 

10 

112 

60 

2 

3 

10 

7 

25 

1 

2 

11 

50 
421 

29 
00 

0 
2 

1 

16 

3 

16 

~3i~ 

1 

8 

3 

^ 

14 

39 

1 

2 

(11.) 

(12.) 

bu. 

pk. 

qt. 

pt. 

3T- 

da. 

h. 

min. 

sec 

23 

3 

7 

1 

25 

300 

19 

54 

35 

34 

2 

0 

1 

21 

40 

12 

40 

24 

42 

3 

5 

0 

3 

112 

14 

15 

17 

51 

1 

4 

1 

6 

10 

11 

45 

59 

23 

0 
3 

3 

4 

0 

0 

1 

1 

1 

1 

1 

11 

57 

109 

11 

37 

16 

13.  If  a  printer  one  day  use  4  bundles  1  ream  15  quires 
20  sheets  of  paper,  the.  next  day  3  bundles  1  ream  10  quires 
10  sheets,  and  the  next  2  bundles  13  sheets,  how  much  does 
he  use  in  the  three  days  ? 

Aiis.    2  bales  1  ream  6  quires  19  sheets. 
11.    A  tailor  used,  in  one  year,  2  gross  5  doz.  10  buttons, 
another  year  3  gross  7  doz.  9,  and  another  year  4  gross  6 
doz.  1 1  ;  how  many  did  he  use  in  the  three  years  ? 

Ans.    10  gross  8  doz.  6. 


COMPOUND   NUMBERS.  185 

15,  A  ship,  leaving  New  York,  sailed  east  the  first  day  o° 
Ao'  50" ;  the  second  day,  4°  50'  10''  ;  the  third,  2°  10'  55  ' ; 
tlie  fourth,  2*^  oO  " ;  how  lar  was  she  then  east  from  the  phice 
of  starting?  Ans.    12°  47' 34''. 

IG.  A  man,  in  digging  a  cellar,  removed  127  cu.  yd.  20  cu. 
ft.  of  earth ;  in  digging  a  drain,  G  cu.  yd.  25  cu.  it. ;  and  in 
digging  a  cistern,  17  cu.  yds.  18  cu.  ft. ;  what  was  the  amount 
of  earth  removed,  and  what  the  cost  at  16  cents  a  cu.  yd.? 

Ans.    152-]- cu.  yds. ;  $24.37^. 

17.  A  farmer  received  80  cents  a  bushel  for  4  loads  of 
corn,  weighing  as  follows  :  2564,  2713,  3000,  and  3109  lbs. ; 
how  much  did  he  receive  for  the  whole?  Ans.    $162.657-j- 

18.  A  druggist  sold  for  medicine,  in  three  years,  at  an  aver- 
age price  of  9  cents  a  gill,  the  following  amounts  of  brandy, 
viz.:  1  bbh4gah  1  pt. ;  30  gal.  2  qt.  1  gi. ;  2  bbh  15  gal.; 
how  much  did  he  receive  for  the  whole  ?        Ans.    §415.17. 

5218.    To  add  denominate  fractions. 

1.  Add  1^  of  a  mile  to  J-  of  a  furlong. 

OPERATION.  Analysis.    We  find  the 

f  mi.  =  6  fur.  26  rd.  11    ft.  "^^^^^  of  each  fraction  in  in- 

1  fLu._^—              13  rd      5-i  ft  tegers  of  less  denominations 

-;;; — -  (213),  and  then  add  their 

Ans.    i  lur.  00           0  values  as  in  comjjound  num- 

Or,  i  fu r.  -^  8  =  jV  nii-  ^""^  ( '^ *  ^  )• 

^V  mi.  +  -I  mi.  =  li  mi.  =  7  fur.      .  ^^''/^"^  .  ^"^^  'f""'?    ^^% 
■^  *  given  tractions  to  fractions  of 

the  same  denomination  (212),  then  add  them,  and  find  the  value 

of  their  sum  in  lower  denominations  (213). 

2.  Add  I  of  a  rod  to  f  of  a  foot.  Ans.    13  ft.  U  in. 

3.  What  is  the  sum  of  |^  of  a  mile,  f  of  a  furlong,  and  f  of 
a  rod  ?  Ans.    7  fur.  27  rd.  8  ft.  3  in. 

4.  What  is  the  sum  of  f  of  a  pound  and  |^  of  a  shilling  ? 

Ans.    13  s.  10  d.  2§  qr. 

5.  What  is  the  sum  of  §  of  a  ton  and  f  of  1  cwt.  ? 

Ans.    12  cwt.  42  1b.  13f  oz. 

Give  explanation  of  the  process  of  addmg  denominate  fractions. 


1S6  SUBTRACTION. 

6.  "What  is  the  sum  of  |  of  a  day  added  to  ^  an  hour  ? 

Alls,    i)  h.  30  mill. 

7.  What  is  the  sum  of  ^  of  a  week,  f  of  a  day,  and  ^  of 
an  hour  ?  Ans.    1  da.  22  h.  15  min. 

8.  Add  f  of  a  hhd.  to  f  of  a  gal. 

9.  What  is  the  sum  of  f  of  a  cwt.,  8f  lb.,  and  Sy^j  oz.  by 
long  ton  table  ?  Ans.    73  lb.  1  oz.  3J-J-  dr. 

10.  What  is  the  sum  of  |  of  a  mile,  f  of  a  yard,  and  ^  of 
a  foot  ? 

11.  Sold  4  village  lots;  the  first  contained  J-  of  ^  of  an 
acre ;  the  second,  G0|^  rods ;  the  third,  f  of  an  acre ;  and  the 
fourth,  §  of  f  of  an  acre  ;  how  much  land  in  the  four  lots  ? 

Ans.    3  E.  2G  P.  120^-^^  sq.  ft. 

12.  A  farmer  sold  three  loads  of  hay ;  the  first  weighed 
1^  T.,  the  second,  1  ^3^.  T.,  and  the  third,  18|  cwt.;  what  was 
the  aggregate  weight  of  the  three  loads  ? 

Ans.   3  T.  5  cwt.  91  lb.  lOS  oz. 


SUBTRACTION. 

*I19.      1.    If  a  druggist  buy  25  gal.  2  qt.  1    pt.  1  gi.  of 
wine,  and  sell  18  gal.  3qt.  1  pt.  2  gi.,  how  much  has  he  left? 

OPERATION.  Analysis.     Writing  the  subtrahend 

gal.     qt.    pt.    gi.         under  the  minuend,  placing  units  of  the 

2'3      2      1      1  same   denomination  under  each  other, 

18      3      0      2  ve  begin   at  the  right  hand,  or  lowest 

^       ~^      o      7^      ^       denomination ;    since   we   cannot    take 

2  gi.  from  1  gi.,  we  add  1  pt.  or  4  gi.  to 
1  gi.,  making  5  gl. ;  and  taking  2  gi.  from  5  gi.,  we  write  the  remain- 
der, 3  gi.,  underneath  the  cokimn  of  gills.  Having  added  1  pt.  or 
4  gi.  to  the  minuend,  we  now  add  1  ])t.  to  the  0  pt.  in  the  subtra- 
liond,  making  1  pt.  ;  and  1  pt.  from  1  pt.  leaves  0  pt.,Mhich  wc  write 
in  the  remainder.  Next,  as  we  cannot  take  3  qt.  from  2  qt.,  wt'  add 
1  gal.  or  4  qt.  to  2  qt.,  making  6qt.,  and  taking  3  qt.  from  0  (p.,  we 
write  the  remainder,  3  qt.,  under  the  denomination  of  quarts.  Add- 
ing 1  gal.  to  18  gal.,  we  subtract  19  gal.  from  2o  gal.,  as  in  simple 

"WTiat  is  subtraction  of  compound  nimibcrs  ?    Give  explanation. 


COMPOUND  NUMBERS. 


187 


numbers,  and  write  the  remainder,  6  gal.,  under  the  column  of  gal- 
lons.    Hence  the  foUoMing 

Rule.  I.  Write  t/ie  subtrahend  under  tlie  nunucnd,  so  that 
units  of  the  same  denomination  shall  stand  under  each  otJier. 

II.  Beginniiuj  at  the  right  hand,  subtract  each  doioiniiiation 
separately,  as  in  simple  numbers. 

III.  If  the  number  of  any  denomination  in  the  subtrahend 
exceed  that  of  the  same  denomination  in  the  minuend,  add  to 
the  number  in  the  minuend  as  many  units  as  make  one  of  tlis 
next  higher  denomination,  and  then  subtract;  in  this  case  add 
1  to  the  next  higher  denomination  of  the  subtrahend  before 
subtracting.  Proceed  in  the  same  manner  icith  each  denomi- 
nation. 


EXAMPLES 

FOR 

PRACTICE. 

lb. 

oz. 

•) 

pwt. 

gT- 

(3.) 
A.   R.   P. 

From 

18 

6 

10 

14 

25   2 

1G.9 

Take 

10 

5 

4 

6 

19   3 

25.14 

Rem. 

8 

1 

6 

8 

5   2 

31.76 

(i-) 

(5.) 

T.   cwt. 

lb. 

5^- 

da. 

h. 

min.   sec. 

14   11 

69f 

38 

187 

16 

45   50 

10   12 

98| 

17 

190 

20 

50   40 

20       361       19       bo       10 

6.  A  Boston  merchant  bought  English  goods  to  the  amount 
of  4327  £  13  s.  7^d.,  and  he  paid  1374  £  10  s.  llf  d. ;  how 
much  did  he  then  owe  ? 

7.  From   300  miles  take  198  mi.  7  fur.  25  rd.  2  yd.    1ft. 

Ans.    101  mi.  14  rd.  2  yd.  2  ft.  8  in. 


10  in 


8.  What  is  the  difference  in  the  longitude  of  two  places, 
one  75=  20'  30"  west,  and  the  other  71°  19'  35  '  west  ?' 

Ans.    4°  55". 

9.  From  10  tb  7  §  4  3  1  9  15  jjr.  take  3ft)8§   2323 


18  or. 


Ans. 


6  1b  11  §  1    3  1  9  17gr. 


Give  the  Rule. 


188  SUBTRACTION. 

10.  The  apparent  periodic  revolution  of  the  sv.n  is  made  in 
36.3  da.  6  h.  9  min.  9  sec.,  and  that  of  the  moon  in  29  da.  12  h. 
44  inin.  3  sec. ;  what  is  the  difference  ? 

Ans.    33.5  da.  1.5  h,  2.5   min.  6  sec. 

11.  A  man,  having  a  hogshead  of  wine,  drank,  on  an  aver- 
age, for  five  years,  including  two  leap  years,  one  gill  of  wine 
a  day ;  how  much  remained  ?         Ans.  5  gal.  3  qt.  1  pt.  1  gi. 

12.  A  section  of  land  containing  6-iO  acres  is  owned  by 
four  men  ;  the  first  owns  196  A.  2  R.  16^  P. ;  the  second,  200 
A.  li  R. ;  the  third,  177  A.  36  P. ;  how  much  does  the  fourth 
own  ?  Ans.    65  A.  3  R.  7.75  P. 

13.  From  a  pile  of  Avood  containing  75^  Cd.  was  sold 
at  one  time  16  Cd.  5  cd.  ft.;  at  another,  24  Cd.  6  cd.  ft.  12 
cu.  ft.  ;  at  another,  27  Cd.  112  cu.  ft.  ;  how  much  remained  in 
the  pile  ?  Ans.    6  Cd.  3  cd.  ft.  4  cu.  ft. 

14.  If  from  a  hogshead  of  molasses  10  gal.  1  qt.  1  pt.  be 
drawn  at  one  time,  15  gal.  1  pt.  at  another,  and  14  gal.  3  qt. 
at  another,  how  much  will  remain  ? 

220.    To  find  the  difference  in  dates. 

1.  What  length  of  time  elapsed  from  the  discovery  of 
America  by  Columbu?,  (jlct.  14,  1492,  to  the  Declaration  of 
Independence,  July  4,  1776? 

FIRST  OPERATION.  ANALYSIS.     "\Ve  place  the  earlier  date 

yr.           mo.       da.  mider  the  later,  writing  first  on  the  left 

^  ^           '           ■*  the  number  of  the  year  from  the  Chris- 

1492        10        14  tian  era,  next  the  number  of  the  month, 

2S3          8        50  counting  January  as  the  first  month,  and 

next  the  number  of  the   day  from  the 

first  day  of  the  month.  Instead  of  the  number  of  the  year,  month, 

and  day,  some  use  the  number  of  years,   months,   and  days   that 

SECOND  OPERATION.  !"'^'^  elapsed  since  the  Christian  era,  thus  : 

,,..            „in_       fjj,/  instead  of  saying  July  is  the  7th  month, 

UJ,'')           6           3  ^^^    ^^V    ^    months    and    .'5    days    have 

■lACj]           n         -I  o  elajisod,  and  instead   of   sayini;:  October 

is  tlie  lOth  montli,  we  say  9  months  and 

283          8        20  13  days  have  elapsed. 


How  is  the  difference  of  dates  found  ? 


COMPOUND   NUMBERS. 


189 


Both  methods  Mill  obtain  the  same  result ;  the  former  is  generally 
used. 

Notes.  1.  When  hours  are  to  be  obtahicd,  -we  reckon  from  12  at 
night,  and  if  minutes  and  seconds,  we  write  them  still  at  the  right  of 
hours. 

2.  In  finding  the  time  between  two  dates,  or  in  computing  interest, 
12  months  are  considered  a  year,  and  30  days  a  month. 

When  the  exact  number  of  days  is  required  for  any  period 
not  exceeding  one  ordinary  year,  it  may  be  readily  found  by 
the  following 

TABLE, 

Slioxcing  the  miynber  of  days  from  any  day  of  one  month  to  the  same  day 
of  any  otlier  mo7ith  within  one  year. 


rilOM  ANY 

TO  THE  SAME  DAY  OF  THE  NEXT. 

DAY  OF 

Jan. 
365 

Feb. 
31 

Mar. 
59 

Apr. 
90 

May. 
120 

151 

July 
181 

Aii^r. 
212 

.«.pt. 
243 

Oct. 
273 

Nov. 
304 

Dec. 

January  .  .  . 

334 

February  .  . 

334 

36.3 

28 

59 

89 

120 

150 

181 

212 

242 

273 

303 

March  .... 

306 

337 

365 

31 

61 

92 

122 

153 

184 

214 

245 

275 

April 

275 

306 

334 

365 

SO 

61 

91 

122 

153 

183 

214 

244 

Mav 

2-15 

276 

304 

335 

365 

31 

61 

92 

123 

153 

184 

214 

June 

214 

245 

273 

304 

334 

365 

30 

61 

92 

122 

153 

183 

July 

184 

215 

243 

274 

304 

335 

36o 

31 

62 

92 

123 

153 

August  .  .  . 

153 

184 

212 

243 

273 

3C4 

334 

365 

31 

61 

92 

122 

September  . 

122 

153 

181 

212 

242 

273 

303 

334 

365 

30 

61 

91 

October.  .  .  . 

92 

123 

151 

182 

212 

243 

273 

304 

335 

365 

31 

61 

November  . 

61 

92 

120 

151 

181 

212 

242 

273 

304 

334 

365 

30 

December.  . 

31 

62 

90 

121 

151 

182 

212 

243 

274 

304 

335 

365 

If  the  days  of  the  different  months  are  not  the  same,  the 
number  of  days  of  difference  should  be  added  vihen  the  earlier 
day  belongs  to  the  moni\\  from  which  we  reckon,  and  subtracted 
when  it  belongs  to  the  month  to  which  we  find  the  time.  If 
the  29th  of  February  is  to  be  included  in  the  time  computed, 
one  day  must  be  added  to  the  result. 

EXAMPLES    FOR   PRACTICE. 

2.  George  Washington  was  born  Feb.  22,  1732,  and  died 
Dec.  14  1799  ;  wdiat  was  his  age  ?   Ans.  67  yr.  9  mo.  22  da. 

How  can  the  nmnber  of  days,  if  less  than  a  year,  be  obtained  ? 


190  SUBTRACTION. 

3.  How  much  time  has  elapsed  since  the  declaration  of 
independence  of  the  United  States  ? 

4.  How  many  years,  months,  and  days  from  your  birthday 
to  this  date  ;  or  wiiat  is  your  age  ? 

5.  How  long  from  the  battle  of  Bunker  Hill,  June  17,  1775, 
to  the  battle  of  Waterloo,  June  18,  1815  ?      Ans.  40  yr.  1  da. 

6.  What  length  of  time  will  elapse  from  20  miimtes  past 
2  o'clock,  P.  M.,  June  24, 1856,  to  10  minutes  before  9  o'clock, 
A.  M.,  January  3,  1861  ?     Ans.  4  yr.  6  mo.  8  da.  18  h.  30  min. 

7.  How  many  days  from  any  day  of  April  to  the  same  day 
of  August  ?  of  December?  of  February? 

8.  How  many  days  from  the  6th  of  November  to  the  15tli 
of  April?  Ans.    160  days. 

9.  How  many  days  from  the  20th  of  August  to  the  15th 
of  the  following  June  ?  Ans.    299  days. 

2S3.    To  subtract  denominate  fractions. 

1.    From  f  of  an  oz.  take  |^  of  a  pwt. 

OPERATION,  Analysis.     We  per- 

|oz.     z=  7  pwt.  12  gr.  form   the    same   reduc- 

z  pwt.  =r  21  o-r.  tions  as  in  addition  of 

denominate      fractions, 

6  pwt.  15  gr.,  Ans.  (218  ),  and  then  sub- 

tract the  less  value  from 
Or,      f  oz.  X  20  r=  Y  pwt.  the  greater. 

¥  —  i  =  ¥-  pwt.  =  6  pwt.  15  gr. 


to' 


2.  What  is  the  difiercnee  between  ^  rod  and  f  of  a  foot  ? 

Ans.    7ft.  6  in. 

3.  From  |-  £  take  f  of  |-  of  a  shilling. 

4.  From  f  of  a  league  lake  -/j  of  a  mile. 

Ans.    1  mi.  2  fur.  16  rd. 

5.  From  S-p^  cwt.  take  1  qr.  2^  lb. 

A)is.    8  cwt.  2  qr.  14  lb.  5  oz.  lOj^dr. 

6.  From  -^  of  a  week  take  ^  of  a  day. 

A71S.    1  da.  4  h.  48  min. 

Give  explanation  of  the  process  of  subtracting  denominate  fractions. 


COMPOUND   NUMBERS.    .  191 

7.  Two  persons,  A  and  B,  start  from  two  places  120  miles 
apart,  and  travel  toward  each  other;  after  A  travels  f ,  and 
1>  #,  of  the  distance,  how  far  are  tliey  apart  ? 

Ans.    41.  mi.  7  fur.  9  rd.  8  ft.  7^  in. 

8.  From  a  cask  of  brandy  containing  9G  gallons,  -^  leaked 
out,  and  f  of  the  remainder  was  sold  ;  how  much  still  remained 
in  the  cask  ?  Ans.    25  gal.  2  qt.  3^  gi. 


MULTIPLICATION. 

223.  1.  A  farmer  has  8  fields,  each  containing  4  A.  2  R. 
27  P. ;  how  much  land  in  all  ? 

OPERATION.  Analysis.     In  8  fields  are  8  times  as  much 

A.     It.      r.  land  as   in  1  field.     We  write  the  muki])lier 

4     2      27  under  the  lowest  denomination   of  the  mul- 

8  tiplicand,   and   proceed   thus;    8  times  27  P- 

~ — ' --"  are    216  P.,    equal  to  5  11.    16  P.;    and   we 

'^'  write  the  16  P.  under  the  number  multiplied. 

Then  8  times  2  R.  are  16  11.,  and  5  11.  added  make  21  R.,  equal  to 
4  A.  1  R. ;  and  we  Avrite  the  1  II.  under  the  number  multiplied. 
Again,  8  times  4  A  are  32  A.,  and  4  A.  added  make  36  A.,  which  we 
write  under  the  same  denomination  in  the  multipUcaud,  and  the 
work  is  done.     Hence, 

Rule.  I.  Write  the  multiplier  under  the  loioest  denomina- 
tion of  the  midtiplicand. 

11.  Multiphj  as  in  simple  members,  and  carry  as  in  addi' 
iion  of  compound  numbers. 

EXAMPLES   FOR   PRACTICE. 

(3.) 

mi.     fur.     rd.         ft. 

9     4     20     13 
6 


(2-) 

bu. 

rk.    qt. 

pt. 

4 

2     5 

1 

2 

9     13     0  57     3       4     12 


Multiplication  of  compovmd  numbers,  how  performed  ?    Rule. 


192 


MULTIPLICATION. 


(4.) 

£.        s.        d. 

5     18     4 

(5.) 

lb.    oz.     pwt.    gr. 

3     4    0     22 

4 

7 

(6.) 

T.        cwt.      lb. 

14     16     48 

oz. 

12 

13°     10'     35" 

11 

9 

8.  In  6  barrels  of  grain,  each  containing  2  bu.  3  pk.  5  qt., 
how  many  bushels?  ^«5-    17  bu.  1  pk.  6  qt. 

9.  If  a  druggist  deal  out  3  lb  4  S  1  5  2  9  16  gr.  of  med- 
icine a  day,  how  much  will  he  deal  out  in  6  days  ? 

10.  If  a  man  travel  29  mi.    3  fur.  30  rd.  loft,  in  1   day, 
how  far  will  he  travel  in  8  days  ? 

11.  If  a  woodchopper  can  cut  3  Cd.  48  cu.  ft.  of  wood  in  1 
day,  how  many  cords  can  he  cut  in  12  days?    Ans.  404-  Cd. 

12.  What  is  the  weight  of  48  loads  of  hay,  each  weighing 
1  T.  3  cwt.  50  lb.  ? 


T, 
1 


cwt. 

3 


OPERATION, 
lb. 
50 
6 


7     1 


00    weight  of  6  loads. 
8 


56       8       00    weight  of -IS  loads. 


Analysis.  "When  the  multi- 
plier is  large,  and  a  composite 
number,  we  may  multiply  by  one 
of  the  factors,  and  that  product 
by  the  other.  Multiplying  the 
weight  of  1  load  by  6,  we  obtain 
the  weight  of  6  loads,  and  the 
weight  of  6  loads  multiplied  by 
8,  gives  the  weight  of  48  loads. 

13.  If  1  acre  of  land  produce  45  bu.  3  pk.  6  qt.  1  pt.  of 
corn,  how  much  will  64  acres  produce?  Ans.    2941  bu. 

14.  How  much  will  120  yards  of  cloth  cost,  at  1  £  9  s.  8^  d. 
per  yard  ? 

15.  If  $80  will  buy  4  A.  3  R.  26  P.  20  ?q.  yd.  3  sq.  ft.  of 
land,  how  much  will  $4800  buy?      Ans.    295  A.  10  pq.yd. 

16.  If  a  load  of  coal  by  the  long  ton  weigh  1  T.  6  cwt.  2  qr. 
26  lb.  10  oz.,  what  will  be  the  weight  of  73  loads  ? 

Ans.   97  T.  11  cwt.  3  qr.  1 1  lb.  10  oz. 


COMPOUND   NUMBERS.  193 

17.  The  sun,  on  an  average,  clianges  his  longitude  59'  8.33" 
per  day;  how  much  will  be  the  change  in  3 Go  days? 

18.  If  1  pt.  3  gi.  of  wine  fill  1  bottle,  how  much  will  be  re- 
quired to  fill  a  great  gross  of  bottles  of  the  same  capacity .? 

DIVISION. 

223.      1.   If  4  acres  of  land  produce  102  bu.  3  pk.  2  qt.  of 
wheat,  how  much  will  1  acre  produce  ? 

OPERATiox.  Analysis.     One  acre  will  produce  ^ 

pt,      bu.      ijU.    qt.    pts.         as  much  as  4  acres.     "Writing  tlie  divi- 

"^  )  ^^'^      ^      ^  sor  ou  tlie  left  of  the  dividend,  we  divide 

2^     2      6     1  102  bu.  by  4,  and  we  obtain  a  quotient  of 

2<j  bu.,  and  a  remainder  of  2  bu.     We 

write  the  25  bu.  under  the  denomination  of  bushels,  and  reduce  the 

2  bu.  to  pecks,  making  8  pk.,  and  the  3  pk.  of  the  dividend  added 

makes  11  pk.     Dividing  11  pk.  by  4,  we  obtain  a  quotient  of  2  pk. 

and  a  remainder  of  3  pk.  ;  writing  the  2  pk.   under  the  order  of 

pecks,  we  next  reduce  3   pk.    to    quarts,    adding  the  2   qt.  of  the 

dividend,  making  26  qt.,  which  divided  by  4  gives  a  quotient  of  G  qt. 

and  a  remainder  of  2  qt.     Writing  the  6  qt.  under  the  order  of 

quarts,  and  reducing  the  remainder,  2  qt.,  to  pints,  we  have  4  pt., 

which  divided  by  4  gives  a  quotient  of  1  pt.,  which  we  write  under 

the  order  of  pints,  and  the  work  is  done. 

2.   A  farmer  put  132  bu.  operation. 

1  pk.  of  apples  into  46  barrels;  '"'•     ''''• 

how  many  bu.  did  he  put  into  4^)1^2     1(2  bu. 
a  barrel ?  ,  '  — 


40 
4 


"\Mien  the  divisor  is  large,  and  .       . 

not  a  composite  number,  we  di-  V  "  P   * 

vide  by  long  division,  as  shown  ^ 

in  the  operation.     From  these  23 

examples  we  derive  the  o 


184(4qt. 

Ans.  2  bu.  3  pk.  4  qt. 


j>  p_      Explain  the  process  of  dividing  compound  nimibera. 


194 


DIVISION, 


Rule.  L  Divide  the  highest  denomination  as  in  simple 
members,  and  each  succeeding  denomination  in  the  same  man- 
ner, if  there  be  no  remainder. 

II.  Jf  there  be  a  remainder  after  dividing  any  denomina- 
tion, reduce  it  to  the  next  lotver  denomination,  adding  in  the 
given  number  of  that  denomination,  if  any,  and  divide  as  be- 
fore. 

III.  The  several  partial  quotients  will  be  the  quotient  re- 
quired. 

Notes.  1.  "Wlien  the  divisor  is  large  and  is  a  composite  number, 
■we  may  shorten  the  work  by  dividing  by  the  factors. 

2.  \Vlien  the  divisor  and  dividend  are  both  compound  nirmbers,  they 
must  both  be  reduced  to  the  same  denomination  before  dividing,  and 
then  the  i3rocess  is  the  same  as  in  simple  numbers. 


£. 

5)25 


(3.) 

s. 

8 


EXAMPLES    FOR    PRACTICE. 


d. 

4 


4)3 


1 

(5.) 

da.        h. 

5      22 


8 


mm. 

00 


T. 

7)  45 

(4.) 

cwt. 

15 

lb. 

25 

6 

10 
(6.) 

75 

))25° 

42' 

40" 

6       17      30 


2        34 


16 


7.  Bought  6  large  silver  spoons,  which  weighed  1 1  oz.  3  pwt.; 
■what  was  the  weight  of  each  spoon  ? 

8.  A  man  traveled  by  railroad  1000  miles  in  one  day; 
what  was  the  average  rate  per  hour  ? 

Ans.    41  mi.  5  fur.  13  rd.  5  ft.  6  in. 
0.    If  a  family  use  10  bbl.  of  flour  in  a  year,  what  is  the 
average  amount  each  day?  Ans.    51b.  5  oz.  14-5T}dr. 

10.  The  aggregate  weight  of  123  hogsheads  of  sugar  is 
57  T.  19  cwt.  42  1b.  14  oz.;  what  is  the  average  weight  per 
hojishead?  Ans.    9  cwt.  42  lb.  10  oz. 

11.  IIow  many  times  are  5  £  10  s.  10  d.  contained  in  537  £ 
10  s.  10  d.?  Ans.   97. 

■  ■■         .      -  ■     -  -^ — '       — ■ 

Give  the  rule.  "SMien  the  divisor  is  a  compo'site  number,  how  may 
we  proceed  ?  ^Y^len  the  divisor  and  dividend  arc  both  compound 
numbers,  how  proceed  ? 


COMPOUND   NUMBERS.  195 

12.  A  cellar  50  ft.  lonpj,  30  ft.  wide,  and  6  ft.  deep  was  ex- 
cavated by  5  men  in  G  davs  ;  how  many  cubic  yards  did  each 
man  excavate  daily?  Ans.    11  cu.  yd.  3  cii.ft. 

13.  If  a  town  !')  miles  square  be  divided  equally  into  150 
farms,  what  will  be  the  size  of  each  farm  ? 

Ans.    lOG  A.  2  R.  26  P.  20  sq.  yd.  1  sq.ft.  72  sq.  in. 

14.  How  many  times  are  4  bu.  3  pk.  2  qt.  contained  in 
336  bu.  3  pk.  4qt.?  Ans.    70. 

15.  A  merchant  tailor  bought  4  pieces  of  cloth,  each  con- 
taining 60  yd.  2.25  qr. ;  after  selling  ^  of  the  whole,  he  made 
up  the  remainder  into  suits  containing  9  yd.  2  qr.  each;  how 
many  suits  did  he  make  ?  Ans.    17. 


LONGITUDE   AND   TBIE. 

S'21.  Every  circle  is  supposed  to  be  divided  into  360 
equal  parts,  called  degrees. 

Since  the  sun  appears  to  pass  from  east  to  west  round  the 
earth,  or  through  360°,  once  in  every  24  hours,  it  will  pass 
through  2^y  of  360^,  or  15"  of  the  distance,  in  1  hour  ;  and  1°  of 
distance  in  J^  of  1  hour,  or  4  minutes;  and  1'  of  distance  in 
•^'g-  of  4  minutes,  or  4  seconds. 

TABLE   OF   LOXGITUDE   AND   TIME. 
3G0°  of  longitude  z=z  24  hours,  or  1  day  of  time. 


5°   " 

l( 

=     1  hour 

>( 

X 

ic     „ 

u 

z=i    4  minutes 

i( 

(( 

1'  " 

ti 

r=     4  seconds 

« 

u 

CASE    I. 

3S«>.    To  find  the  tlifforencc  of  time  between  two 
places,  when  their  longitudes  are  given. 

1.    Tlie  longitude  of  Boston  is  71'  3',  and  of  Chicago  87^ 
30' ;  what  is  the  difference  of  time  between  these  two  places  ? 


Explain  how  distance  is  measured  by  time.      Repeat  the  table  of 
longitude  and  time.      Case  I  is  what  ? 


OPEKATION. 

87° 

30' 

71 

3' 

16° 

27' 

4 

19 g  LONGITUDE   AND   TDIE. 

Analysis.  By  subtraction  of 
compound  numbers  v.e  first  find 
the  difference  of  longitude  be- 
tween the  two  places,  which  is 
16°  27'.  Since  1°  of  longitude 
makes  a  difference  of  4  minutes 

1  ,      I      '.       TZ"  A  of  time,  and   1'  of  longitude  a 

1  h.  o  rain.  48  sec,  Ans.  ,.^.       '       ,  ,  ,      p    • 

dmerence  or  4  seconds  oi  tunc, 

we  multiply  16°  27',  the  difference  in  longitude,  by  4,  and  we  obtain 

the  difference  of  time  in  minutes   and  seconds,  wliich,  reduced  to 

higher  denominations,  gives  1  h.   5  min.  48  sec,  the  difference  in 

time.     Hence  the 

Rule.     3Tultiply  the  difference  of  longitude  in  degrees  and 

minutes  by  4,  and  tJie  product  will  be  the  difference  of  time  in 

minutes  and  seconds,  wliich  may  he  reduced  to  hours. 

Note.  If  one  place  be  in  cast,  and  the  other  in  west  longitude,  the 
difference  cf  longitude  is  found  by  adding  them,  and  if  the  sum  be 
greater  than  180°,  it  must  be  subtracted  fiom  360°. 

EXAMPLES    FOR    PRACTICE. 

2.  New  York  is  74'  1'  and  Cincinnati  84^  24'  west  longi- 
tude;  what  is  the  difference  of  time  ?    Ans.  41  min.  32  sec. 

3.  The  Cape  of  Good  Hope  is  18^  28'  cast,  and  the  Sand- 
wich Islands  155' west  longitude;  what  is  the  difference  of 
time  ?  Ans.    1 1  h.  33  min.  52  sec. 

4.  Washington  is  77°  1'  west,  and  St.  Petersburg  30' 
19'  east  longitude  ;  what  is  their  difference  of  time  ? 

Ans.    7  h.  9  rain.  20  sec. 

5.  If  Pekin  is  118'  east,  and  San  Fiancisco  122'  west 
longitude,  what  is  their  diflerence  of  time  ? 

G.    If  a  message  be  sent  by  telegraph  without  any  loss  of 

time,  at   12  M.  from  London,  0°  0'  longitude,  to  Washington, 

77°  V  Avest,  what  is   the  time  of  its  receipt  at  Washington? 

Note.  Since  the  sun  ajipcars  to  move  from  on«t  to  west,  when  it  is 
exactly  12  o'clock  at  one  place,  it  will  be  pant  12  o'clock  at  all  places 
east,  and  before  12  at  all  places  west.  Hence,  knowina;  the  difference 
of  time  between  two  places,  and  the  exact  time  at  one  of  them,  the 
exact  time  at  the  other  will  be  foimd  by  addinfj  their  difference  to  the 
given  time,  if  it  be  east,  and  by  siiblracfing  if  it  be  west. 

Ans.    6  h.  51  min.  b(j  sec,  A.  M. 
Give  explanation.    Kule. 


COMPOUND   NUMBERS.  197 

7.  A  steamer  arrives  at  Halifax,  G3°  36'  west,  at  4  o'clock, 
P.  M. ;  the  fact  is  telegraphed  to  St.  Louis,  90^  15'  west, 
without  loss  of  time;  what  is  the  time  of  its  receipt  at  St. 
Louis  ?  Ans.    2  h.  13  mill.  24  sec,  P.  M. 

8.  If,  at  a  presidential  election,  the  voting-  begin  at  sunrise 
and  end  at  sunset,  how  much  sooner  will  the  polls  open  and 
close  at  Eastport,  Me.,  G7^  west,  than  at  Astoria,  Oregon,  124° 
west  ?  Ans.    3  h.  48  min. 

'J.  When  it  was  1  o'clock,  A.  M.,  on  the  first  day  of  Jan- 
uary, 18o9,  at  Bangor,  Me.,  68°  47'  west,  what  was  the 
time  at  the  city  of  Mexico,  99°  5'  west? 

Ans.    Dec.  31,  1858,  58  min.  48  sec.  past  10,  P.  M. 

CASE    II. 

2*J<I.  To  find  the  difference  of  longitude  between 
two  places,  when  the  difference  of  time  is  known. 

1.  If  the  difference  of  time  between  New  York  and  Cincin- 
nati be  41  min.  32  sec,  what  is  the  difference  of  lonjiitude  ? 

OPERATION.  Analysis.     Since  4  minutes  of  time 

mill.      sec.  make  a  difl'erence  of  1°  of  longitude,  and 

4 )  41        32  '              4  seconds  of  time,  a  difference  of  1'  of 

7~      7^  longitude,  there  will  be  J-  as  many  de- 

"^  '  ■          grecs  of  longitude  as  there  are  minutes 

of  time,  and  ^  as  many  minutes  of  longitude  as  there  are  seconds  of 

time.     Hence, 

Rule.  Iteduce  the  difference  of  time  to  minutes  and  sec- 
onds, and  then  divide  hy  4;  the  quotient  will  he  the  difference 
in  longitude,  in  degrees  and  minutes. 

2.  "What  is  the  ditverence  of  longitude  between  the  Cape 
of  Good  Hope  and  the  Sandwich  Islands,  if  the  difference  of 
lime  lie  11  h.  33  min.  52  sec?  Ans.    173    28'. 

3.  What  is  the  difference  of  loniritude  between  Washington 
and  St.  Petersburg,  if  their  difference  of  time  be  7  h.  9  min. 
20  sec?  Ans.    107°  20'. 

Case  II  is  what  ?     Give  explanation.     Rule. 


193  DUODECIMALS. 

4.  When  it  is  half  past  4,  P.  M.,  at  St.  Petersburg,  30^  19' 
east,  it  is  32  miii.  36  sec.  past  8,  A.  M.,  at  New  Orleans,  west ; 
what  is  the  diiference  of  longitude?  Ans.    ll'J    21' 

5.  The  longitude  of  New  York  is  74'  1'  west.  A  sea  cap- 
tain leaving  that  port  for  Canton,  with  New  York  time,  finds 
that  his  chronometer  constantly  loses  time.  AVhat  is  his  longi- 
tude when  it  has  lost  4  hours  ?  8  h.  40  min.  ?  13  h.  25  min.  ? 

Ans.    14^  r  west;  55'  59'  east;   127'  14'  east. 
G.    When  the  days  are  of  equal  length,  and  it  is  noon  on 
the  1st  meridian,  on  what  meridian  is  it  then  sunrise?  sun- 
set ?  midnight  ?    Ans.  90'  west ;  90'  east ;  180  east  or  west. 


DUODECIMALS. 

fB^7,  Duodecimals  are  the  divisions  and  subdivisions  of 
a  unit,  resulting  from  continually  dividing  by  12,  as  1,  yV?  t¥3'' 
•jy^g,  &c.  In  practice,  duodecimals  are  applied  to  the  meas- 
urement of  extension,  the  foot  being  taken  as  the  unit. 

If  the  foot  be  divided  into  12  equal  parts,  the  parts  are 
called  inches,  or  primes  ;  the  inches  divided  by  12  give  sec- 
onds;  the  seconds  divided  by  12  give  thirds;  the  thirds  di- 
vided by  12  give  fourths;  and  so  on. 

From  these  divisions  of  a  foot  it  follows  that 

1'     (inch  or  prime) is    ^\    of  a  foot. 

1''    (second)  or  jV  of  yV "  yJ-y   of  a  foot. 

V"  (third)  or  tV  of  tV  of  tV'  •  •  "  rrVff  of  a  foot,  ice. 

TABLE. 

12  fourths,  marked  (""),  make  1  third marked  V" 

12  thirds  "      i  second,  "        \" 

12  seronds  «      ]  prime,  or  inch,  "        V 

12  primes,  or  inclies,  "      1  foot,  "        ft. 

SCAl.K  — uniforndy  12. 

The  marks  ',  ", '",  "",  arc  called  indices. 

\^Tiat  arc  diioflccimals  ?  To  what  applied  ?  Explain  ths  di-s-isiona 
of  tlie  foot.     Repeat  the  table. 


I 


COMPOUND   NUMBERS.  199 

Note.  Duodecimals  are  really  common  fi-actions,  and  can  ahva3-s 
be  treated  as  such  ;  but  usually  their  denom  Jiators  are  not  expressed, 
and  they  are  treated  as  compound  numbers. 

Addition  and  Subtraction  op  Ddodkcimals. 

S528.  We  add  and  subtract  duodecimals  the  same  as  other 
compound  numbers. 

EXAMPLES. 

1.  Add  13  ft.  4'  8",  10  ft.  6'  7",  145  ft.  9'  11". 

Ans.    169  ft.  9'  2". 

2.  Add  179  ft.  11'  4",  245  ft.  1'  4",  3  ft.  9'  9". 

Ans.    428  ft.  10'  5". 

3.  From  25  ft.  6'  3"  take  14  ft.  9'  8".    Ans.  10  ft.  8'  7". 

4.  From  a  board  15  ft.  T  G"  in  length,  3  ft.  8'  11"  Avere 
sawed  off;  what  was  the  lenglh  of  the  piece  left? 

Ans.   11  ft.  10'  7". 

Multiplication  of  Duodecijials. 

S^O.  Length  multiplied  by  breadth  gives  surface,  and 
surface  multiplied  by  thickness  gives  solid  contents  (898). 

1.  How  many  square  feet  in  a  board  11  feet  8  inches  long 
and  2  feet  7  inches  wide? 

OPERATION.  Analysis.    We  first  multiply  by  the  7'. 

1 1  ft.     8'  7  twelfths  times  8  twelfths  equals  56  one 

2  7'  hundred     forty- fourths,     which    equals    4 

twelfths  and  8  one  hundred  forty-fourths. 


6  ft.      9'      8"  We  Avrite  the  8  144ths — marked  with  two 

23  4'  indices  —  to  the  ri":ht,  and  add  the  4  12ths 


"Oft       T      8"  ^°  ^^^^  next  pi'oduct.      7'  times  11   equals 

77',  which  added  to  4'  equals  81',  equal  to 
6  feet  and  9'.  We  write  the  9'  under  the 
inches,  or  12ths,  and  the  6  under  the  feet,  or  units.  2  times  8' 
equals  16',  or  1  foot  and  4'.  We  Avritc  the  4'  under  tlie  9',  and 
add  the  1  foot  to  the  next  product.  2  times  1 1  feet  are  22  feet,  and 
1  foot  added  make  23  feet,  which  we  write  under  the  6  feet.  Add- 
How  are  duodecimals  added  and  subtracted  ?  Give  analysis  of  ex- 
ample 1.  , 


200  DUODECIMALS. 

ing  these  partial  products,  and  ■we  have  30  ft.  1'  and  8"  for  the 
entire  product. 

It  will  be  seen  from  the  above  that  the  number  of  indices  to  every 
product  of  any  two  factors  is  equal  to  the  sum  of  the  indices  of  those 
factors  ;  thus  7'  X  B'  ^  56"  ;  4"  X  o'"  =z  20'"".     Hence  the 

Rule.  I.  Write  the  several  terms  of  the  multiplier  under 
the  corresfondtng  terms  of  the  multiplicand. 

II.  Multiply  each  term  of  the  midtiplicand  hy  each  term  of 
Hie  multiplier,  beginning  with  the  lowest  term  in  each,  and  call 
the  product  of  any  two  denominations  the  denomination  denoted 
hy  the  sum  of  their  indices,  carrying  1  for  every  12. 

III.  Add  the  partial  products,  carrying  1  for  every  12; 
their  sum  will  he  the  required  answer. 

EXAMPLES    FOR    PRACTICE. 

2.  How  many  square  feet  in  a  board  13  ft.  9'  long  and  \V 
wide?  Ans.    12ft.  7'  3". 

3.  IIow  many  square  feet  in  a  stock  of  4  boards,  each  1 1  ft. 
9'  long  and  1  ft.  3'  wide  ?  Ans.    58  ft.  9'. 

4.  How  many  square  yards  of  plastering  on  the  walls  of  a 
room  12  ft,  11'  square,  and  9  ft.  3''  high,  allowing  for  two  win- 
dows and  one  door,  each  G  ft.  2'  high  and  2  ft.  4'  wide  ? 

Ans.    48  sq.  yd.  2  ft.  9'. 

5.  How  many  solid  feet  in  a  mow  of  hay  30  ft.  4'  long, 
25  ft.  6'  wide,  and  12  ft.  5'  high  ?  Ans.    9G04  ft.  3'  6". 

6.  How  many  cords  in  a  pile  of  wood  18  ft.  G'  long,  12  ft. 
■wide,  and  5  ft.  6'  high  ?  ^ns.    9  cords  G9  ft. 

7.  How  many  cubic  yards  of  enrth  must  be  removed  in 
digging  a  cellar  36  ft.  10'  long,  22  ft.  3'  wide,  and  5  ft.  2'  deep  ? 

Am.    156cu.yd.  22  ft.  3'  1". 

8.  What  would  it  cost  to  plaster  a  Avail  32  ft.  8'  long  and 
9  ft.  higli,  at  17  cents  per  square  yard  ?  Ans.    $5.55^. 

9.  How  many  yards  of  carpeting,  27'  wide,  will  be  re- 
quired to  cover  a  floor  48  ft.  long  and  33  ft.  9'  wide  ? 

Ans.    240  yards. 


Give  the  nde. 


COMPOUND   NUMBERS.  201 

Division  of  Duodecimals. 
*2S0.    1.    A  flagstone,  3  ft.  9'  wide,  has  a  surface  of  20  fl 
11'  3"  ;  what  is  its  length? 

oPERATiox.  Analysis.     We  divide 

3  ft.  9'  )  20  ft.  1 1'     3"  (  5  ft.  7'.         the  surface   by  the  width 

]  g         9'  to  obtain  the  length.     The 

divisor  is  something  more 

2         2'     3'  tj^jjj-^   3  fj^  ,j,jj  ^^  obtain 

2  2      o  the  first  quotient  figure,  v.e 


consider  how  many  times 
3  ft.  and  something  more  is  contained  in  nearly  21ft.  (20  ft.  11'); 
we  estimate  it  to  be  o  times,  and  multiplying  the  divisor  by  this 
quotient  figure,  we  have  18  ft.  9',  which,  subtracted  from  20  ft.  11', 
leaves  2  ft.  2',  to  which  we  bring  down  3'',  the  last  term  of  the  divi. 
dend.  We  next  seek  how  many  times  the  divisor  is  contained  in 
this  remainder,  and  find  by  trial  the  quotient  7' ;  multiplying  the 
divisor  by  this  figure,  we  obtain  2  ft.  2'  o",  and  there  is  no  remain- 
der.    Hence  the 

Rule.     I.    Wrife  the  divisor  on  the  left  hand  of  the  dividend, 
as  in  simple  7i umbers. 

II.  Find  tlie  frst  term  of  the  quotient  either  hy  dividing  the 
frst  term  of  the  dividend  hy  the  first  term  of  the  divisor,  or  hy 
dividing  the  first  two  terms  of  the  dividend  by  the  first  two 
terms  of  the  divisor  ;  multiply  the  divisor  by  this  term  of  the 
quotient  subtract  the  product  from  the  corresponding  terms  of 
the  dividend,  and  to  the  remainder  bring  down  anotlter  term  of- 
the  dividend. 

III.  Proceed  in  like  manner  till  there  is  no  remainder,  or 
till  a  quotient  has  been  obtained  sufiicienily  exact. 

EXAMPLES    FOR    PRACTICE. 

2.  Divide  44  ft.  5'  4'  by  IGft.  8',  Ans.   2  ft.  8'. 

3.  The   square   contents  of  a   walk  arc  184  ft.  3',  and  tho 
length  is  40  ft.  11'  4"  ;  what  is  the  width?       Ans.    4ft.  6'. 

.4.  A  blanket  whore  squan  contents  are  14  ft.  6',  is  to  be 
lined  with  cloth  2  ft.  7'  wide  ;  how  much  in  length  will  be  re- 
quired? 

Give  analysis  of  example  1.     Rule. 
9* 


2<,2 


PROmSCUOUS   EXAMPLES. 


5.  A  block  of  granite  contains  G4  ft.  2'  5"  ;  its  widJi  is 
2  ft.  6',  and  its  tliickness  3  ft.  7'  ;  what  is  its  lengtli  ? 

Note.  Since  the  solid  content.!  are  the  product  of  the  three  dimen- 
sions, we  divide  the  sclid  contents  by  any  two  dnuensions  or  by  their 
product,  to  obtain  the  other  dimension. 

Ans.    7  ft.  2'. 

PROMISCUOUS    EXAMPLES. 

1.  In  115200  grains  Troy,  how  many  pounds? 

2.  In  365  da.  5  h.  48  min.  46  sec,  how  many  secondt:  ? 

Ans.   3155G92G. 

3.  A  man  wishes  to  ship  15 GO  bushek  of  potatoes  in  bar- 
rels containing  3  bu.  1  pk.  each ;  how  many  barrels  will  be 
required  ?  -^ns.    480. 

4.  Reduce  295218  inches  to  miles. 

5.  lleduce  456575  grains  to  pounds,  apothecaries'  weight. 

Ans.    79  1b  3  i  13  1  9  15  gr. 

6.  How  many  sheets  in  3  reams  of  paper  ? 

7.  AVhat  is  the  value  of  4  piles  of  wood,  each  20  ft.  long,  G  ft. 
wide,  and  10  ft.  high,  at  $3.25  per  cord?      Ans.    $121.87^. 

8.  How  many  bottles,  each  holding  1  qt.  1  gi.,  can  be  filled 
from  a  barrel  of  cider  ?  Ans.    112. 

9.  At  $20.40  per  sq.  rd.  for  land,  what  will  \^p,  the  cost  of  a 
village  lot  8^  rd.  long,  and  ^  rd.  wide  ?         Ans.    $980.10. 

10.  Divide  259  A.  1  R.  10  P.  of  land  into  3G  equal  lots. 

Ans.    7  A.  321  P.  ^ 

11.  How  many  times  can  a  box  holding  4  bu.  3  pk.  2  qt.  be 
filled  from  336bu.  3  pk.  4qt.?  Ans.    70. 

12.  What  io  the  value  of  .875  of  a  gallon  ? 

13.  What  part  of  a  mile  is  2  fur.  36  rd,  2  yd.  ?    Ans.  i\. 

1 4.  What  part  of  2  days  is  13  h.  26  min.  24  sec.  ? 

15.  From  2G  A.  2  R.  of  land,  5  A.  3  R.  were  sold  ;  what 
part  of  the  whole  piece  remained  unsold?  Ans.    -^^y^- 

1  G.  Wh;it  is  the  difference  between  ^  of  a  pound  sterling 
and  5;|-  pence?  Ans.    11  s.  6^- d. 

1 7.  What  is  the  sum  of  f  of  a  yard,  |  of  a  foot,  and  |  of 
an  inch?  Ans.    7  inches. 


PROMISCUOUS   EXAMPLES.  '  203 

18.  Reduce  G  cwt.  Iqr.  7  lb.  of  coal  to  the  decimal  of  a 
long  ton.  Ans.    .lGoG25. 

19.  Benjamin  Franklin  was  born  Jan.  18,  170G,  and  George 
Washington  Feb.  22,1732-,  bow  much  older  was  Franklin 
than  Washington  ?  "  Ans.    2G  yr.  1  mo.  4  da. 

20.  The  longitude  of  Boston  is  71*^  4'  west,  and  that  of 
Chicago  87°  30'  west;  when  it  is  12  M.  at  Boston,  what  is  the 
time  in  Chicago?  Ans.    10  h.  54  min.  IG  sec.  A.  M. 

21.  If  the  ditfcrence  of  time  between  New  York  and  New 
Orleans  be  1  h.  4  sec,  what  is  the  difference  in  longitude  ? 

Ans.     15°  1'. 

22.  Add  §  of  a  mile,  ^  of  a  furlong,  and  ^%  of  a  rod  to- 
gether. Ans.    5  fur,  33  rd.  8  ft.  3  in. 

23.  If  a  bushel  of  barley  cost  $.80,  what  will  20  bu.  3  pk. 
6qt.  cost?  Ans.   $16.75. 

24.  What  is  the  value  of  c875  of  a  gross  ?    Ans.    10^  doz. 

25.  How  many  acres  in  a  field  56^  rods  long,  and  24.6 
rods  wide  ?  Ans.   8  A.  2  K.  29.9  P. 

26.  How  many  perches  of  masonry  in  the  wall  of  a  cellar 
which  is  20  feet  square  on  the  inside,  8  feet  high,  and  1^  feet 
in  thickness  ?  ,  Ans.    44.6-(-. 

27.  A,  B,  and  C  rent  a  farm,  and  agree  to  work  it  n[)on 
shin-PS ;  they  raise  640  bu.  3  pk.  of  grain,  which  they  divide 
as  follows  :  one  fourth  is  given  tor  tlie  rent ;  of  the  remainder 
A  takes  1  JJ-  bu.  more  than  one  third,  after  which  B  takes  one 
Jialf  of  the  remainder  less  7  bush<ils,  and  C  lias  what  is  left ; 
how  much  is  C's  share  ?  Ans,    161  bu.  3  pk.  G  qt. 

28.  What  is  the  value  in  Troy  weight  of  Id  lb.  8  oz.  1 1.4  dr. 
avoirdupois  weight?      Ans.   16  lb.  5  oz,  10  pwt.  11.7  -\- gr. 

29.  If  154  bu.  1  pk.  6  qt.  cost  $173.74,  how  much  will  1.5 
bushels  cost?  Ans.    $1,687+. 

SO.    What  is  the  value  of  .0125  of  a  ton?     Ans.    25  lbs. 
3L    What  fraction  of  3i)ushel3  is  y^  of  2  bu.  3  pk.  ? 

jlns.    xiT" 
32.    How  many  wine  gallons  in  a  water  tank  4  feet  long, 
3^  feet  wide,  and  1  ft.  8  in.  deep?  Ans.    174^. 


20-i  PROMISCUOUS   EXAMPLES. 

S3.  How  many  bushels  will  a  bin  contain  that  is  7^-  feet 
square,  and  G  ft.  8  in.  deep?  Ans.    301.339  -|-  bu. 

34.  How  much  must  be  paid  for  lathing  and  plastering 
overhead  a  room  3G  feet  long  and  20  feet  wide,  at  '2{}  cents  a 
square  yard  ? 

35.  How  many  shingles  will  it  take  to  cover  the  roof  of  a 
building  46  feet  long,  each  of  the  two  sides  of  the  roof  being 
20  feet  wide,  allowing  each  shingle  to  be  4  inches  wide,  and 
to  lie  5  inches  to  the  weather?  Ans.    1324S. 

36.  John  Young  was  born  at  a  quarter  before  4  o'clock,  A. 
M.,  Sept.  4,  1836;  what  will  be  his  age  at  half  past  6  o'clock, 
P.  M.,  April  20,  1864  ?      Ans.  27  yr.  7  mo.  1 6  da.  14  h.  45  min. 

37.  How  many  cubic  yards  of  earth  were  removed  in  dig- 
ging a  cellar  28  ft.  9'  Lng,  22  ft.  8'  wide,  and  7  ft.  6'  deep  ? 

Ans.    181 -'j  cu.  yd. 

38.  What  will  30  bu.  54  lb.  of  wheat  cost,  at  $1.37J-  per 
bushel?  Ans.    $42.4875. 

39.  How  many  square  yards  of  carpeting  will  it  take  to 
cover  a  floor  24  ft.  8'  long  and  18  ft.  6'  wide  ?     Ans.    SO^S. 

40.  Wiiat  is  the  cost  of  54  bu.  8  lb.  of  barley,  at  84  cents 
per  bushel  ?  Ans.    $45.50. 

41.  YvHiat  is  the  depth  of  a  lot  that  has  120  feet  front,  and 
contains  18720  square  feet? 

42.  How  many  steps  of  30  inches  each  must  a  person 
take  in  walking  21  miles? 

43.  How  long  will  it  require  one  of  the  heavenly  bodies  to. 
move  through  a  quadrant,  if  it  move  at  the  rate  of  3'  12" 
per  minute  ?  Aiis.    1  da.  4  h.  7  min.  30  sec. 

44.  How  many  times  will  a  wheel,  9  ft.  2  in.  in  circum- 
ference, turn  round  in  going  65  miles  ? 

45.  If  a  man  buy  10  bushels  of  chestnuts,  at  $5.00  per 
bushel,  dry  measure,  and  sell  the  same  at  22  cents  per  quart, 
licjuid  measure,  how  much  is  his  g^n?  Ans.    $31.92. 

46.  What  will  it  cost  to  build  a  wall  210  i'vct  long,  6  feet 
high,  and  3  feet  thick,  at  $3.25  per  1000  bricks,  each  brick 
being  8  inches  long,  4  inches  wide,  and  2  inches  thick  ? 

A}is.    $379.08. 


PERCENTAGE.  205 


PERCEJ^TAGE. 


SSI .  Per  cent,  is  a  term  derived  from  the  Latin  words  per 
centum,  and  signifies  hy  the  hundred,  or  hundredths,  tliat  is,  a  cer- 
tain number  of  parts  of  eacli  one  hundred  parts,  of  whatever  de- 
nomination. Thus,  by  5  per  cent,  is  meant  5  cents  of  every  100 
cents,  $5  of  every  $100,  5  bushels  of  every  100  busliels,  &c. 
Therefore,  5  per  cent,  equals  5  hundredths  rrr  .05  rr:  -^^-^  zrz  ^'j. 
8  per  cent,  equals  8  hundredths  zzz  .08  =  y§-^  z=z  ■^^. 

S£SS.  Percsntage  is  such  a  part  of  a  number  as  is  indi- 
cated by  the  per  cent. 

S«sS.  The'  Ease  of  percentage  is  the  number  on  which 
the  percentage  is  computed. 

SS-J:.  Since  per  cent,  is  any  number  of  hundredths,  it  is 
usually  expressed  in  the  form  of  a  decimal  j  but  it  may  be 
expressed  either  as  a  decimal  or  a  common  fraction,  as  in  tho 


following 

TABLE. 

Decimals. 

Commou  Fractions.    Lowest  Terms. 

1 

per  cent. 

— 

.01 

— 

T"0T 

:= 

lOT 

2 

per  cent. 

(( 

.02 

(( 

O 

TTO 

(1 

tV 

4 

- 

per  cent. 

(( 

.04 

i( 

Too 

(( 

^V 

i                     ' 

per  cent. 

« 

.05 

« 

5 

ToT 

(( 

■io 

1                     6 

per  cent. 

« 

.06 

(( 

Toir 

(( 

t\ 

7 

per  cent. 

a 

.07 

« 

7 

T"?ro 

a 

ToT 

8 

per  cent. 

it 

.08 

it 

Tn7 

a 

O 

2t 

10 

per  cent. 

11 

.10 

<( 

iVo 

(f 

1 
1  0 

16 

per  cent. 

(( 

.16 

<( 

(( 

A 

20 

per  cent. 

(( 

.20 

(( 

2  0 
Tno" 

it 

i 

25 

per  cent. 

(( 

.25 

i( 

25 

1  no 

it 

i 

60 

per  cent. 

(< 

.50 

(1 

1  no 

it 

i 

100 

per  cent. 

<( 

1.00 

tt 

IILQ. 

1  II  0 

(< 

1 

125 

per  cent. 

(( 

1.25 

(( 

1  2  5 

10  0 

a 

5 

4 

1 

per  cent. 

(1 

.005 

(( 

Tfrrrg- 

a 

TnU 

1 

per  cent. 

a 

.0075 

« 

To  M  n  0 

it 

400 

121 

per  cent. 

11 

.125 

(( 

125 

1  Ti  0  0 

ii 

i 

161 

per  cent. 

li 

.1625 

(( 

1625 

ToooT 

tt 

u 

"What  is  meant  by  per  cent.  ?  From  what  is  the  term  derived  ? 
"What  is  percentage  ?  \Miat  is  the  base  of  percentage  ?  How  is  per 
cent,  expressed? 


20G 


PERCENTAGE. 


EXAMPLES    FOR   PRACTICE, 

1.  Express  decimally  3  per  cent. ;  6  per  cent. ;  9  per  cent. ; 
14  per  cent. ;  24  per  cent. ;  40  per  cent. ;  112^  per  cent. ;  150 
per  cent. 

2.  Express  decimally  G^  per  cent. ;  8f  per  cent. ;  33^  per 
cent.;  7^  per  cent.;  lOf  per  cent.;  9|  per  cent. ;  103J- [ler 
cent. ;  225  per  cent. 

3.  Express  decimally  |  per  cent. ;  f  per  cent. ;  f  per  cent.; 
f  per  cent. ;  |  per  cent. ;  1^  per  cent. ;  2|-  per  cent. ;  4^  per 
cent.;  5f  per  cent.;  7^  per  cent.;  12 1  per  cent.;  25 1  per 
cent. 

4.  Express  in  the  form  of  common  fractions,  in  their  lowest 
terms,  6  per  cent. ;  8  per  cent. ;  12  per  cent. ;  14^  per  cent. ; 
18^'  per  cent. ;  2l|  per  cent.;  31  J- per  cent. ;  371  per  cent. ; 
40^  per  cent.;  112  per  cent.;  225  per  cent. 


CASE   I. 


65 


S5.    To  find  the  percentage  of  any  number. 

1.    A  man,  having  $125,  lost  4  per  cent,  of  it;  how  many 
dollars  did  he  lose  ? 


OPERATION. 

$125 
.04 


$5.00 


Analysis.  Since  4  per  cent,  is  y^=:  .04,  he  lost 
.04  of  $125,  or  8125  X  .04  =  $5.  Or,  4  per  cent, 
is  ifo  =  -^5^,  and  gV  of  $125  :=  $5.     Hence    the 


Rule.  3fuUipIy  the  given  number  or  quantity  hy  the  rate 
•per  cput.  expressed  decimally,  and  ])oiiit  off  as  in  decimals.    Or, 

7\dce  such  a  part  of  the  given  number  as  the  number  ex- 
pressing the  rate  is  part  of  100. 


EXAMPLES    EOi:    PPwVCTICE. 

2.    AVhat  is  G  per  cent,  of  $320  ?  Ans.    $19.20. 


3. 


What  is  8  per  cent,  of  $327.25  ? 


Ans.    $2G.18. 


Case  I  is  what  ?     Give  explanation,     llule. 


PERCENTAGE.  207 

4.  ^Yhat  is  7^  per  cent,  of  $56.75  ?         A71S.    $4.11 /g. 

5.  AVliat  is  12i^  per  cent,  ot"  2450  pounds? 

Ans.    306.25  pounds. 

6.  What  is  6a  per  cent,  of  10072  bushels? 

Aiis.    1287.36  bushels. 

7.  What  is  33^  per  cent,  of  846  gallons  ? 

Ans.    282  gallons. 

8.  What  is  Of  per  cent,  of  275  miles?      Ans.  26.05  miles- 
0.    What  is  14  per  cent,  of  450  sheep  ? 

10.  What  is  50  per  cent,  of  1240  men  ? 

11.  What  is  105  per  cent,  of  $5760  ?  Ans.    $6048. 

12.  What  is  175  per  cent,  of  $12067  ? 

13.  What  is  25  per  cent,  of  |  ? 

2o  per  cent,  equals  ^^^\  =  \,  and  |  X  }  ^=  -^o,  Ans. 

14.  What  is  15  per  cent,  of  |  ?  Ans.    j'5- 

15.  What  is  2^  per  cent,  of  65  ?  Ans.    ^. 

16.  What  is  33^  per  cent,  of  Y^(j  ?  A)is.   ^^. 

17.  What  is  84  per  cent,  of  7^-?  Ans.    Qj%. 

18.  Find  f  per  cent,  of  $40.80  Ans.    $.306. 
10.  Find  If  per  cent,  of  $15.60  Ans.    $.26. 

20.  A  farmer,  having  760  sheep,  kept  25  per  cent,  of  them, 
and  sold  the  remainder ;  how  many  did  he  sell  ? 

21.  A  man  has  a  capital  of  $24500;  he  invests  18  per 
cent,  of  it  in  bank  stock,  30  per  cent,  of  it  in  railroad  stocks, 
and  the  remainder  in  bonds  and  mortoanjes  ;  how  much  does 
lie  invest  in  bonds  and  mortirages?  Ans.   $12740. 

22.  A  speculator  bought  1576  barrels  of  ajiples,  and  upon 
opening  them  he  found  12^  per  cent,  of  them  spoiled;  how 
many  barrels  did  he  lose  ? 

23.  Two  men  engaged  in  trade,  each  Avith  $2760.  One  of 
them  gained  33^  per  cent,  of  his  capital,  and  the  other  gained 
75  per  ceat. ;   how  much  more  did  the  one  gain  than  the  other  ? 

Ans.    $1150. 

24.  A  man,  owning  |-  of  an  iron  foundery,  sold  35  per  cent, 
of  his  share ;  what  part  of  the  whole  did  he  sell,  and  Avhat 
part  did  he  still  own  ?  Ans.    He  still  owned  ^l 


203 


PERCENTAGE. 


25.  A  owed  B  $575.40  ;  he  paid  at  one  time  40  per  cent, 
of  the  debt;  afterward  he  paid  25  per  cent,  of  the  remainder; 
and  at  another  time  124-  per  cent,  of  what  he  owed  after  ihe 
second  payment;  how  nmch  of  the  debt  did  he  still  owe  ? 


A/is. 


§226.50^ 


CASE   11. 

S2S.  To  find  what  per  cent,  one  number  is  of  an- 
otlier. 

1.  A  man,  having  $125,  lost  $5;  what  per  cent,  of  his 
money  did  he  lose  ? 

OPERATION. 

.04  =r  4  per  cent. 
Or, 
=  vjV"  ^^  •'^^  ^^  "^  P^^'  cent. 


5-^125 


—5- 


Analysis.  We  multi- 
ply the  base  by  the  rate 
per  cent,  to  obtain  the 
percentage  (tj;i . 5) ;  con- 
versely, we  divide  the  per- 
centage by  the  base  to  obtain  the  rate  per  cent.  Or,  since  $120  is 
100  per  cent,  of  his  money,  $5  is  y|-j>  equal  to  -^^  of  100  per  cent. 
which  is  4  per  cent.     Hence  the 

Rule.  Divide  the  percentage  hj  the  base,  and  the  quotient 
will  be  the  rate  per  cent,  expressed  decimally.     Or, 

Take  such  a  part  of  100  as  the  percentage  is  part  of  the 
base. 


EXAMPLES    FOR    rUACTICE. 

2.    What  per  cent,  of  $  150  is  $00  ? 


Ans.    20. 
Ans.    12.i. 


3.  What  per  cent,  of  $1400  is  $175? 

4.  What  per  cent,  of  $750  is  $1  G5  ? 

5.  Wiiat  per  cent,  of  $2-10  is  $13.20?  Ans.    5^-. 
r..    Wliat  per  cent,  of  $2  is  15  cents  ? 

7.    What  per  cent,  of  G  bushels  1  peck  is  4  bushels  2  pecks 
G  quarts  ?  Ans.    75  per  cent. 

,     8.    What  per  cent,  of  15  pounds    is    5    pounds    10    ounces 
avoirdupois  weight  ?  Ans.    373- per  cent. 

9.   What  per  cent,  of  250  head  of  cattle  is  40  head  ? 

Cose  II  is  what  ?     Give  explanation.     Rule. 


1 


PERCENTAGE.  209 

10.  From  a  hogshead  of  sugar  containing  7G0  pounds,  100 
pounds  were  sold  at  one  time,  and  90  pounds  at  another;  what 
per  cent,  of  the  whole  was  sold  ? 

11.  A  man,  having  GOO  acres  of  land,  sold  ^  of  it  at  one 
time,  and  ^  of  the  remainder  at  another  time;  what  per  cent, 
remaaied  unsold  ?  Ans.    50  per  cent. 

CASE    III. 

SS7,  To  find  a  number  when  a  certain  per  cent,  of 
it  is  given. 

1.  A  man  lost  $5,  which  was  4  per  cent,  of  all  the  money 
he  had ;  how  much  had  he  at  first  ? 

OPERATION.  Analysis.    We  are  here  required  to 

$5  -f-  .04:  =  $125.  find  the  base,  of  which   $5  is  the  per- 

Or,  centage.     Now,  percentage  equals  base 

-^  X  100  zr:  Si  25  multiphed  by  the  rate  per  cent. ;  con- 

*  versely,  base  equals  percentage  divided 

by  rate  per  cent.     Or,  $5  is  4  per  cent,  of  all  he  had  ;  |  of  $5,  or  -I, 

equals  1  per  cent,  of  all   he  had,  and  100  times  |  equals  100  per 

cent.,  or  all  he  had.     Hence  the 

Rule.  Divide  the  percentage  ly  the  rate  per  cent,,  ex- 
pressed decimalhj,  and  the  quotient  loill  he  the  lase,  or  number 
required.     Or, 

Take  as  many  times  100  as  the  percentage  is  times  the  rate 
per  cent, 

EXAMPLES    FOR   PRACTICE. 

2.  1 6  is  8  per  cent,  of  what  number  ?  Ans.   200. 

3.  42  is  7  per  cent,  of  what  number  ? 

4.  75  is  12^  per  cent,  of  what  number  ?  Ans.    600. 

5.  33  is  2f  per  cent,  of  what  number?  Ans.    1200. 

6.  $281.25  is  374  per  cent,  of  what  sum  of  money  ? 

Ans.    $750. 

7.  A  farmer  sold  50  sheep,  which  was  20  per  cent,  of  his 
whole  flock ;  how  many  sheep  had  he  at  first  ? 

Case  ni  is  what  ?     Give  explanation.     Rule. 


210  PERCENTAGE. 

8.  I  loaned  a  man  a  certain  sum  of  money ;  at  one  time 
he  paid  me  $59.75,  which  ^as  12^^  per  cent,  of  the  whole  buni 
loaned  to  him  ;  how  much  did  I  loan  him  ? 

9.  A  mercliant  invested  $975  in  dry  goods,  wliieli  was  15 
per  cent,  of  his  entire  capital ;  what  was  the  amount  of  his 
capital?  -^ns.    $G5U0. 

10.  If  a  man,  owning  40  per  cent,  of  an  iron  fpundery,  sell 
25  per  cent,  of  his  share  for  $12-40.50,  what  is  the  value  of 
the  whole  foundery  ?  ^'is.    $12405. 

11.  A  merchant  pays  $75  a  month  for  clerk  hire,  which  is 
25  per  cent,  of  his  entire  pro.Hts ;  how^  much  are  his  profits  for 
one  year,  after  paying  his  clerk  hire  .'*  Ans.   $2700. 

12.  A  produce  buyer,  having  a  quantity   of  corn,  bought 
2000  bushels  more,  and  he  found  that  this  purchase  was  40 
per  cent,  of  his  whole  stock ;    how  much  had   he  before  he. 
bou!?ht  this  last  lot  ?  Ans.   3OU0  bushels. 


COMMISSION   AND   BROKERAGE. 

flS8,  An  Agent,  Factor,  or  Broker,  is  a  person  who  trans- 
acts business  for  another,  or  buys  and  sells  money,  stocks, 
notes,  &c. 

!S30.  Commission  is  the  percentage,  or  compensation 
allowed  an  agent,  factor,  or  commission  merchant,  for  buying 
and  selling  goods  or  pi'oduce,  collecting  money,  and  transact- 
ing other  business. 

S'^IO.  Brokerage  is  the  fee,  or  allowance  paid  to  a  broker 
or  dealer  in  money,  stocks,  or  bills  of  exchange,  for  making 
exchanges  of  money,  buying  and  selling  stocks,  negotiating 
bills  of  exchange,  or  transacting  otlier  like  business. 

Note.  The  ratos  of  commi'ision  and  brokcra2;Garc  not  recrnlatod  hy 
law,  but  are  usually  reckoned  at  a  certain  per  cent.  ui)on  the  money 
employed  m  the  transaction. 

Define  an  agent,  factor,  or  broker.  "\Miat  Is  meant  by  commission  ? 
Brokerage  ? 


* 


COMMISSIOx\   AND   BROKERAGE.  211 

CASE  I, 

241.  To  find  the  commission  or  brokerage  on  any 
smn  of  money. 

1.  A  commission  merchant  sells  butter  and  cheese  to  the 
amount  of  §loJ:0  ;  what  is  his  commission  at  o  per  cent.  ? 

OPEKATION.  Analysis. 

1540  X  -05  =1  $;77,  Alts.  Since   the   com- 

Or,  t5  J  =  5-V  ,  and  g-'j  X  1540  =  $77.  mh^lon  on  $1  is 

5  cents  or  .Oo  of 
a  dollar,  on  81-340  it  is  $1540  X  .05:=  $77.  Or,  since  5  per  cent 
^^  loT^^-sV  o^  the  sum  received,  the  commission  is  ^L  of  $1540 
r=  $77.     Hence  the 

RuLK.  Multiply  the  given  sum  hj  the  rate  per  cent,  ex- 
pressed decimally,  and  the  result  ivill  he  Uie  commission  or  bro- 
kerage.    Or, 

2'akc  such  a  part  of  the  given  sum  as  the  number  expressing 
thz  per  cent,  is  part  o/lOO. 

EXAMPLES    FOn    PUACTICE. 

2.  A  commission  merchant  sells  goods  to  the  amount  of 
$6756  ;  what  is  his  commission  at  2  per  cent.  ?   Ans.  $iy5.12. 

3.  What  commission  must  be  paid  for  collecting  $17380, 
at  3J-  per  cent.  ?  Am.    $608.30. 

4.  An  agent  in  Chicago  purchased  4700  bushels  of  wheat, 
at  75  cents  a  bushel ;  what  Avas  his  commission  at  14  per  cent, 
on  the  jiurchase  money? 

5.  A  broker  in  New  York  exchanged  $25875  on  the  Suf- 
folk Bank,  Boston,  at  ^  per  cent. ;  how  much  brokerage  did 
he  receive?  Ans.    $64.6875. 

6.  An  auctioneer  sold  at  auction  a  house  for  $3284,  and 
the  furniture  for  $2170.50  ;  what  did  his  fees  amount  to  at 
2^  per  cent.  ? 

7.  A  broker  negotiates  a  bill  of  excliange  of  S2890  for  ^ 
per  cent,  commission  ;  how  much  is  his  brokerage  ? 

Ans.   $23.12. 

Case  I  is  what  ?     Give  explanation.     Rule. 


212  PERCENTAGE. 

8.  An  agent  buys  for  a  manufacturing  company  2G750 
pounds  of  wool,  at  32  cents  a  pound,  and  receives  a  commis- 
sion of  2f  per  cent. ;  what  amount  does  he  receive? 

Arts.    $235.40. 

0.  If  I  sell  400  bales  of  cotton,  each  weighing  570  {)ounds, 
at  9  cents  a  pound,  and  receive  a  commission  of  2}  per  cent., 
how  much  do  I  make  by  the  transaction  ?       Ans.    $401.70. 

10.  A  commission  merchant  in  New  Orleans  sells  450  bar- 
rels of  flour  at  $7.G0  a  barrel;  38  firkins  of  butter,  each  con- 
taining 56  pounds,  at  25  cents  a  pound  ;  and  105  cheeses,  each 
weighing  48  pounds,  at  9  cents  a  pound  ;  how  much  is  his 
commission  for  selling,  at  5i  per  cent.?        Atis.    $242,308. 

11.  A  lawyer  collected  a  note  of  §950,  and  charged  GJ-  per 
cent,  commission ;  what  was  his  fee,  and  what  the^  sum  to  be 
remitted?  Ans.    Fee,  $G1.75  ;  remitted,  $888.25. 

12.  An  insurance  agent's  fees  are  G  per  cent,  on  all  sums 
received  for  the  com])any,  and  4  per  cent,  additional  on  all 
sums  remaining,  at  the  end  of  the  year,  after  tlie  losses-  are 
paid  ;  he  receives,  during  the  year,  $30456.50,  and  ])ays  losses 
to  the  amount  of  $19814.15;  how  much  commission  does  he 
receive  during  the  year  ?  Ans.    $2253.084. 

CASE    II. 

24:3.  To  find  the  commission  or  brokerage,  when 
it  is  to  bo  deducted  from  the  given  sum,  and  the  bal- 
ance invested. 

1.  A  merchant  sends  his  agent  $1260  with  which  to  buy 
merchandise,  after  deducting  his  commission  of  5  per  cent. ; 
■what  is  the  sum  invested,  and  liow  much  is  the  commission  ? 

OPEnATION. 
$1260  -4-  1.05  =r  $1200,  invested. 
$1260  — $1200  =  $60,  commission. 

Or,  i?,?y  -j-  T^.T  =   U  ;    ^^260  H-  U  =  $1200,   invested; 
And   $1260  —  $1200  =  $60,  ron>nus.inn. 


Case  U  is  what  ?     Give  o.xiilanation.     Rule. 


COMMISSION   AND    BROKERAGE.  213 

Analysis.  Since  the  commission  is  5  per  cent.,  the  agent  must 
receive  $1.05  for  every  $1  he  expends  ;  he  can  invest  as  many- 
dollars  as  $1.0J  is  contained  times  in  $1260,  which  is  $1200;  and 
the  difl'erence  between  the  given  sum  and  the  sum  invested  is  his 
commission. 

Or,  the  money  expended  is  |gj]  of  itself,  the  commission  is  ^|-q-  of 
this  sum,  and  the  commission  added  to  the  sum  expended  is  ii|  of 
the  whole  sum.  Since  $1260  is  1§|  z=z  fi,$126U  ^U  =  '^l^OO, 
the  sum  expended;  and  $1260  —  $1200=;  $60  the  commission. 
Hence  the 

Kt'LE.  I.  Divide  the  given  amount  by  1  increased  ly  the  rate 
per  cent,  of  commission,  and  the  quotient  is  tlie  sum  invested. 

II.  Subtract  the  investment  from  the  given  amount,  and  the 
remainder  is  the  commission, 

EXAMPLES    FOR    PKACTICE. 

2.  A  man  sends  $3246.20  to  his  agent  in  Boston,  request- 
ing him  to  lay  it  out  in  shoes,  after  deducting  his  commissioa 
of  2  per  cent;  how  much  is  his  commission  ?    Ans.  $G3.65. 

3.  What  amount  of  stock  can  be  bought  for  $9G82,  and  al- 
low 3  per  cent,  brokerage  ?  Ans.    $9400. 

4.  A  flour  merchant  sent  $10246.50  to  his  agent  at  Chica- 
go, to  invest  in  flour,  after  deducting  his  commission  of  3^ 
per  cent. ;  how  many  barrels  of  flour  could  he  buy  at  $5..50 
per  barrel?  Ans.    1800  barrels. 

5.  An"  agent  receives  a  remittance  of  $4908,  with  which  to 
purchase  grain,  at  a  commission  of  4^-  per  cent. ;  what  will  be 
the  amount  of  the  purchase  ? 

6.  Remitted  $603.75  to  my  agent  in  New  York,  for  the  pur- 
chase of  merchandise,  agent's  commission  being  5  per  cent. ; 
what  amount  of  broadcloth  at  $5  per  yard  should  I  receive? 

Ans.    115  yds. 

7.  A  commission  merchant  receives  $9376.158,  with  or- 
ders to  purchase  grain  ;  his  co.Timission  is  3  per  cent.,  and  he 
charges  1^-  per  cent,  additional  for  guaranteeing  its  delivery  at 
a  specified  time ;  how  much  will  he  pay  out,  and  what  are 
his  fees  ?  Ans.  Fees,  $403,758. 


214  PERCENTAGE. 

8.  A  real  estate  broker,  whose  stated  commission  is  1^ 
per  cent.,  receives  $13842.07,  to  be  used  in  the  purchase  of 
city  lots  ;  how  much  does  he  invest,  and  what  is  his  commis- 
sion ?  Alls.    $1.") 604  invested;   $238.07  commission. 

9.  A  broker  received  $10050,  to  be  invested  in  stocks  after 
deducting  ^  per  cent,  for  brokerage ;  Avhat  amount  of  stock 
did  he  purchase  ? 

STOCKS. 

9^3.  A  Corporation  is  a  body  authorized  by  a  general 
law,  or  by  a  special  charter,  to  transact  business  as  a  single 
indivi(hiai. 

iM^^^,  A  Charter  is  the  legal  act  of  incorporation,  and  de- 
fines the  powers  and  obligations  of  the  incorporated  body. 

^^4:e3.  A  Firm  is  the  name  under  which  an  unincorporated 
company  transacts  business. 

S46.  Capital  or  Stock  is  the  property  or  labor  of  an  indi- 
vidual, corporation,  company,  or  firm  ;  it  receives  differen'.; 
names,  as  Bank  Stock,  Railroad  Stock,  Government  Stock,  &c. 

24:7.  A  Share  is  one  of  the  equal  parts  into  which  the 
stock  is  divided. 

S^!^.    Stockholders  are  the  owners  cf  the  shares. 

S49.  The  Nominal  or  Par  Value  of  stock  is  its  first  cos*, 
or  original  valuation. 

Note.  The  original  value  of  a  share  varies  in  different  companies. 
A  share  of  bank,  insurance,  railroad,  or  like  stock  is  usually  |ilOO. 

Sr5^.  Stock  is  At  Par  when  it  sells  for  its  first  cost,  or 
original  valuation  ; 

imi ,  Above  Par,  at  a  premium,  or  in  advance,  wlien  it 
foils  ibr  more  tlian  its  original  cost;  and 

^vi*2.  Below  Par,  or  at  a  discount,  wlicn  it  sells  for  less 
than  its  original  cost. 

Define  a  corporation.  A  charter.  A  firm.  Capital  or  stock.  Shares. 
Stockholders.    I'ar  value.    At  par.    Above  par.    Ik'low  par. 


STOCKS.  215 

2*TS.  The  Market  or  Real  Value  of  stock  is  what  it  will 
bring  per  share  in  money. 

•254.  A  Dividend  is  a  sum  paid  to  stockholders  from  the 
profits  of  t lie  business  of  the  company. 

S#5«5.  An  Assessment  is  a  sum  i-equired  of  stockholders  to 
meet  the  losses  or  expenses  of  the  business  of  the  comiiaiiy. 

'^9i&.  Premium  or  advance,  and  discount  on  stock,  divi- 
dends, and  assessments,  are  computed  at  a  certain  per  cent. 
upon  the  original  value  of  the  shares  of  the  stock. 

CASE    I. 

2tl7.  To  find  the  value  of  stock  when  at  an  ad- 
vance, or  at  a  discount.  - 

1.  What  will  $3240  of  bank  stock  cost,  at  8  per  cent,  ad- 
vance ? 

OPEKATION.  Analysis.    Since 

$1  -f-  .08  =  $1.08  $1  of  the  stock  at 

$3240  X  $1.08  —   $3409.20,  Ans.  P^r  value  will  cost 

$1  plus  the  premi- 
um, or  $1.08,  $3240  of  the  same  stock  will  cost  3240  X  $1.08  = 
83499.20.  If  the  stock  were  8  per  cent,  below  par,  $1  minus  the 
discount,  or  $1.00  —  $.08  =:  $.92,  M-ould  show  what  $1  of  the  stock 
would  cost.     Hence  the 

Rule.  M}dtipJy  the  par  value  of  the  stock  hy  the  nianhcr 
indicating  the  price  of  $1  of  the  same  stock,  and  the  product 
ivill  he  the  real  value. 

Note.     In  all  examples  relating  to  stocks,  $100  is  considered  the 
par  value  of  a  share  of  stock,  unless  otherwise  stated. 

KXAMPLES    FOR    PRACTICE. 

2.  If  the  stock  of  an  insurance  company  sell  at  5  per  cent. 
below  par,  what  will  $1200  of  the  stock  cost?   Ans.  $1140. 

3.  What  is  the  market  value  of  35  shares  of  New  York 
Central  Railroad  stock,  at  15  per  cent,  below  par  ? 

iSIarkot  value.  A  dividend.  An  assessment.  Case  I  is  what  ?  Give 
explanation.     Rule. 


216  .       PERCENTAGE. 

4.  What  must  be  paid  for  48  shares  of  Panama  RaHroad 
•«?tock,  at  a  premium  of  5^  per  cent.,  if  the  par  value  be  $150 

per  share  ?  Atis.    $7596. 

5.  What  costs  $53G4  stock  in  the  Minnesota  copper  mines, 
at  9  per  cent,  above  par  ? 

6.  A  man  purchased  $G275  stock  in  the  Pennsylvania  Coal 
Company,  and  sold  the  same  at  a  discount  of  12  per  cent.; 
what  was  his  loss  ?  Ans.    $753. 

7.  Wliat  must  be  paid  for  125  shares  of  United  States 
stock,  at  4|-  per  cent,  premium,  the  par  value  being  SlOOO  per 
share?  Ans.    $130937.50. 

8.  Bought  42  shares  of  Illinois  Central  Railroad  stock,  at 
14  per  cent,  discount,  and  sold  the  same  at  an  advance  of  12^ 
percent.;  how  much  did  I  gain  .'*  Ans,    $1113. 

9.  What  is  the  market  value  of  175  shares  of  stock  in  the 
Suffolk  Bank,  at  |-  per  cent,  advance  ?        Ans.    $17031.25. 

10.  Bought  75  shares  of  stock  in  the  Bank  of  New  Orleans, 
of  $50  each,  at  3  per  cent,  discount,  and  sold  it  at  2|-  per  cent, 
advance;  what  was  my  gain  ?  Ans.    $190,875. 

11.  B  exchanged  28  shares  of  bank  stock,  of  $50  each, 
v/orth  7  per  cent,  premium,  for  25  shares  of  railroad  stock,  of 
$100  each,  at  12^  per  cent,  discount,  and  paid  the  difference 
in  cash;  how  much  cash  did  he  pay  ?  Ans.    $G89.50. 

CASE    II. 

S58.  To  find  how  miicli  stock  may  be  purchased  for 
a  given  sura. 

1.  How  many  shares  of  bank  stock,  at  3  per  cent,  advance, 
may  be  bought  for  $5150? 

OPEKATION.  Analysis.    Since  the  stock 

$5150  -^  1.03  r=  $5000  =  is  at  3  per  cent,  advanco,  $1 

50  shares    Ans.  of  stock  at  par  will  cost  $1.03; 

and  if  we  divide  $5150,  the 
whole  sum  to  he  expended,  by  $1.03,  the  cost  of  $1  of  stock,  the 
quotient  must  be  the  amount  of  stock  purchased.     Hence  the 

Case  II  is  what  ?     Give  explanation. 


PROFIT  AND  LOSS.  217 

EuLE.  Divide  the  given  stun  hy  the  cost  of  $1  of  stock, 
and  the  quotient  will  be  the  nominal  amowit  of  stock  purchased. 

2.  How  many  shares  of  railroad  stock,  at  5  per  cent,  ad- 
vance, can  be  purchased  for  $6300  ?  Ans.    60  shares. 

3.  I  invested  $6187.50,  in  Ocean  Telegraph  stock,  at  10 
per  cent,  discount ;  how  much  stock  did  I  purchase  ? 

Ans.    $6875. 

4.  I  sent  my  agent  $53500  to  be  invested  in  Illinois  Cen- 
tral Railroad  stock,  which  sold  at  7  per  cent,  advance  ;  what 
amount  did  he  purchase  ?  Ans.    $50000. 

5.  Sold  50  shares  of  stock  in  a  Pittsburg  ferry  company, 
at  8  per  cent,  discount,  and  received  $1150;  what  is  the  par 
value  of  1  share  ?  Ans.   $25. 

PROFIT  ANB   LOSS. 

SSO.  Profit  and  Loss  are  commercial  terms,  used  to  ex- 
press the  gain  or  loss  in  business  transactions,  which  is  usually 
reckoned  at  a  certain  per  cent,  on  the  prime  oiv  first  cost  of 
articles. 

CASE   I. 

26®.  To  find  the  amount  of  profit  or  loss,  "when  the 
cost  and  the  gain  or  loss  per  cent,  are  given. 

1.  A  man  bought  a  horse  for  $135,  and  afterward  sold  him 
for  20  per  cent,  more  than  he  gave ;  how  much  did  he  gain  ? 

OPEKATION.  Analysis.      Since    $1 

$135  X  .20  r=  $27,  Alls.  g^ins  20  cents,  or  20  per 

dr.    20   i.<2:iq^vi  «;97         *^"'^t.,  $135  will  gain  $135 

'1""        °  ^  X-20  =  $27.    Or,  smce  20 

per  cent,   equals  ^VV  =  h  the  whole  gain  will  be  ^  of  the  cost. 
Hence  the  following 

Rule.  Ilidtiply  the  cost  hy  the  rate  per  cent,  expressed 
decimally.     Or, 

Take  such  part  of  the  cost  as  the  rate  per  cent,  is  part  o/lOO. 

Rule.     "S\Tiat  is  meant  by  profit  and  loss  ?     Case  I  is  what  ?     Give 
explanation.     Rule. 
E.P  10 


218  '  PERCENTAGE. 


EXAMPLES    FOR    PRACTICE. 


2.  A  grocer  bought  a  hogshead  of  sugar  for  $84.80,  and  sold 
it  at  12^  per  cent,  j^rofit ;  what  was  his  gain  ? 

3.  A  miller  bought  500  bushels  of  wheat  at  $1.15  a  bushel, 
and  he  sold  the  flour  at  16|  per  cent,  advance  on  the  cost  of 
the  wheat;  what  was  his  gain?  Ans.  $1)5.83 A. 

4.  Bought  76  cords  of  wood  at  $3. 62 J-  a  cord,  and  sold  it 
so  as  to  gain  26  per  cent. ;  what  did  I  make  ? 

5.  A  hatter  bought  40  hats  at  $1.75  apiece,  and  sold  them 
at  a  loss  of  14f  per  cent. ;  what  was  his  whole  loss  ? 

6.  A  grocer  bought  3  barrels  of  sugar,  each  containing  230 
pounds,  at  8^  cents  a  pound,  and  sold  it  at  18jSj-  per  cent,  profit ; 
what  was  his  whole  gain,  and  what  the  selling  price  per  pound  ? 

Ans.  Whole  gain,  $10.35  ;  price  per  pound,  9f  cents. 

7.  A  sloop,  freighted  with  3840  bushels  of  corn,  encoun- 
tered a  storm,  when  it  was  found  necessary  to  throw  37^  per 
cent,  of  her  cargo  overboard ;  what  was  the  loss,  at  62^-  cents 
a  bushel  ?  Ans.    $900  loss. 

8.  A  genfleman  bought  a  store  and  contents  for  $4720 ;  he 
sold  the  same  for  12^  per  cent,  less  than  he  gave,  and  then 
lost  15  per  cent,  of  the  selling  price  in  bad  debts;  Avhat  was  his 
entire  loss?  ^??s.  $1209.50. 

9.  A  man  commenced  business  with  $3000  capital ;  the 
first  year  he  gained  2 2 J-  per  cent.,  which  he  added  to  his  capi- 
tal ;  the  second  year  he  gained  30  per  cent,  on  the  whole  sum, 
which  gain  he  also  put  into  his  business ;  the  third  year  he 
lost  16f  per  cent,  of  his  entire  capital ;  how  much  did  he  make 
in  the  3  years  ?  A7is.    $981.25. 

CASE    II. 

SGI.  To  find  the  gain  or  loss  per  cent.,  when  the 
cost  and  selling  price  are  given. 

1.  Bought  wool  at  32  cents  a  pound,  and  sold  it  for  40  cents 
a  pound ;  what  per  cent,  was  gained  ? 

Case  n  is  what  ?     Give  explanation.     Rule. 


PROFIT   AND   LOSS.  219 

OPERATION. 

40  —  32  =  8  ;  8  -H  32  =  ^8^  =  .25,  Ans. 
Or,  40  —  32  z=  8 ;  8  -^  32  =  ^^  =  ^;  "i  X  100  =  25  per  cent. 

Analysis.  Since  the  gain  on  32  cents  is  40  —  32  =  8  cents,  the 
whole  gain  is  -r^^  =  \  of  the  i^urchase  money ;  and  \  reduced  to  a 
decimal  is  2.3  hundredths,  equal  to  2o  jier  cent.  Or,  if  the  gain  were 
equal  to  the  purchase  money,  it  would  be  100  jjer  cent. ;  but  since 
the  gain  is  -.f^  r=  1  of  the  purchase  money,  it  will  be  \  of  100  per 
cent.,  equal  to  2j  per  cent.     Hence  the  following 

Rule.     3Iake  the  difference  hetween  the  purchase  and  selling 

prices  the  numerator,  and  the  purchase  price  the  denominator  ; 

reduce  to  a  decimal,  and  the  result  will  be  the  per  cent.     Or, 

Take  such  a  part  of  100  as  the  gain  or  loss  is  part  of  the 
purchase  price. 

EXAJrrLES    FOR    PRACTICE. 

2.  A  man  bought  a  pair  of  horses  for  $275,  and  sold  them 
for  $330  ;  what  per  cent,  did  be  gain  ?     Ans.    20  per  cent. 

3.  If  a  merchant  buy  cloth  at  $.60  a  yard,  and  sell  it  for 
$.75  a  yard,  what  does  he  gain  per  cent.  ? 

4.  A  speculator  bought  108  barrels  of  flour  at  $4,624  a 
barrel,  and  sold  it  so  as  to  gain  $114,881;  Avhat  per  cent, 
profit  did  he  make?  ^„^^.    93  per  cent. 

5.  Bought  sugar  at  8  cents  a  pound,  and  sold  it  for  9^  cents 
a  pound ;  what  per  cent,  was  gained  ? 

6.  A  drover  bought  150  head  of  cattle  for  $42  per  head, 
and  sold  them  for  $5400  ;  what  Avas  his  loss  per  cent.  ? 

Ans.    141  per  cent. 

7.  If  I  sell  for  $15  what  cost  me  $25,  what  do  I  lose  per 
°^'''^-  ^  A71S.    40  per  cent. 

8.  Bought  paper  at  $2  per  ream,  and  sold  it  at  25  cents 
a  quire;  what  was  the  gain  per  cent.  ?    Ans.  150  per  cent. 

0.  If  I  sell  ^  of  an  article  for  f  of  its  cost,  wdiat  is  gained 
P^^  cent.  ?  Ans.    50  per  cent. 

10.    If  4  of  an  article  be  sold  for  what  ^  of  it  cost,  what  is 
'  tbe  loss  per  cent.  ?  Ans.  37^  per  cent. 


220  PERCENTAGE. 

11.  If  I  sell  3  pecks  of  clover-seed  for  what  one  bushel 
cost  me,  what  per  cent,  do  I  gain  ?  A7is.    33^  per  cent. 

12.  A,  having  a  debt  against  B,  agreed  to  take  $.87^-  on 
the  dollar ;  what  per  cent,  did  A  lose  ? 

13.  A  grocer  bought  7  cwt.  20  lb.  of  sugar,  at  7  cents  a 
pound,  and  sold  3  cwt.  42  lb.  at  8  cents,  and  the  remainder  at 
8^  cents  ;  what  was  his  gain  per  cent.  ?  Ans.  IS^^g  per  cent. 

14.  Bought  2  hogsheads  of  wine,  at  $1.25  a  gallon,  and 
sold  the  same  at  $1.G0  ;  what  was  the  whole  gain,  and  what 
the  gain  per  cent.  ?  Aiis.    Gain  2S  per  cent. 

15.  A  grain  dealer  bought  corn  at  $.55  a  bushel  and  sold 
it  at  $.66,  and  wheat  for  $1.10,  and  sold  it  for  $1.37^;  upon 
which  did  he  make  the  greater  per  cent.  ? 

Ans.   5  per  cent.,  upon  the  wheat. 

CASE   III. 

S6*3.  To  find  the  selling  price,  when  the  cost  and 
the  gain  or  loss  per  cent,  are  given. 

1.  Bought  a  horse  for  $136  ;  for  how  much  must  he  be  sold 
to  gain  25  per  cent.  ? 

OPERATION.  Analysis.    Since  $1  of  cost 

$1  -f-  .25  =z  $1.25.  sells  for  $1.25,  $136  of  cost  will 

$1.25  X  136  z=z  $170,  Ans.     sell  for  136  times  81.2.5,  which 

n     ioo  _1_  _?.i.  —  a->f.  —  5  equals  -Si 70,  the  selhng  price. 

'^i,  1  o'i  -f  1^0  —  1  ''^~^'  Or,  since  the  cost  is  \^,  and 

$136  X  i  =  $170,  A71S.        ^Yic  gain  ^2J_,  the  selling  price 

Mill  be  |3lrr:i  of  the  cost,  or 
I  of  $136  =  $170.  If  the  horse  had  been  sold  at  a  loss  of  25  per 
cent.,  then  $1  of  cost  would  have  sold  for  $1  minus  .25,  or  $.75, 
&c.     Hence, 

Rule.  3Iultiphj  $1  increased  hy  the  gain  or  diminished  hij 
the  loss  per  cent,  hy  the  number  denoting  the  cost.     Or, 

Take  such  a  part  of  the  cost  as  is  equal  to  \%%  increased  or 
diminished  hy  the  gain  or  loss  per  cent. 


Case  in  is  what  ?     Give  explanation.     Rule. 


PROFIT   AND   LOSS.  221 


EXAMPLES    FOR    rRACTICE. 


2.  If  124-  liundred  weight  of  sugar  cost  $140,  how  must 
it  be  sold  per  pound  to  gain  25  per  cent.  ?      Ans.    14  cents. 

3.  Bought  a  hogshead  of  molasses  for  30  cents  a  gallon, 
and  paid  IGf  per  cent,  on  the  prime  cost,  for  freight  and  cart- 
age ;  how  much  must  it  sell  for,  per  gallon,  lo  gain  334-  per 
cent,  on  the  whole  cost?  Jns.    $.4G§. 

4.  For  what  price  must  I  sell  coifee  that  cost  10^  cents  a 
pound,  to  gain  17^  per  cent.? 

5.  If  I  am  compelled  to  sell  damaged  goods  at  a  loss  of  15 
per  cent.,  how  should  I  mark  goods  that  cost  me  $.02^  ? 
$1.20?  $3.87^?  Ans.    $.53^;  $1.02;  $3.29|." 

6.  A  man,  wishing  to  raise  some  money,  offers  his  house 
and  lot,  which  cost  him  $3240,  for  18  per  cent,  less  than  cost; 
"A^'hat  is  the  price  ? 

7.  C  bought  a  farm  of  120  acres,  at  $28  an  acre,  paid 
$480  for  fencing,  and  then  sold  it  for  12^  per  cent,  advance 
on  the  Avhole  cost;  what  was  his  whole  gain,  and  wdiat  did  he 
receive  an  acre  ?  Ans.    $480  gain  ;  $36  an  acre. 

8.  Bought  a  cask  of  brandy,  containing  52  gallons,  at 
$2. GO  per  gallon  ;  if  7  gallons  leak  out,  how  must  the  remain- 
der be  sold  per  gallon,  to  gain  37^  per  cent,  on  the  cost  of  the 
whole?  Ans.    $4.13^. 

9.  A  merchant  bought  15  pieces  of  broadcloth,  each  piece 
containing  23^  yards,  for  $840,  and  sold  it  so  as  to  gain  18|- 
per  cent, ;  how  much  did  ho  receive  a  yard  ? 

CASE    IV. 

^63.  To  find  the  cost,  when  the  selhng  price  and 
the  gain  or  loss  per  cent,  are  given. 

1.  A  merchant  sold  cloth  for  $4.80  a  yard,  and  by  so  doing 
made  33-J  j^er  cent. ;  how  much  did  it  cost  ? 

OPERATION. 

1  +  .331  =  1.33^  ;  $4.80  ^  1.33i  =  $3.G0,  Ans. 
Or,  $4.80  =  f  of  the  cost ;  $4.80  -^f  =  $3.60. 


Case  IV  is  what  ? 


222  PERCENTAGE. 

Analysis.  Since  the  gain  is  33^  per  cent,  of  the  cost,  $1  of  the 
cost,  increased  by  !33^  per  cent.,  will  be  what  $1  of  cost  sold  for  : 
therefore  there  will  be  as  many  dollars  of  cost,  as  l.oo^-  is  con- 
tained times  in  $4.80,  or  $3.60.  Or,  since  he  gained  33^  per  cent. 
=  1  of  the  cost,  $4.80  is  f  of  the  cost ;  $4.80-:-  |-  =  $3.60, 

Note.  If  the  rate  per  cent,  be  loss,  we  subtract  it  from  1,  instead 
of  adding  it.     Hence  the  foUowuig 

Rule.  Divide  the  selling  price  hy  1  increased  hij  the  gain 
or  diminished  hy  the  loss  per  cent.,  expressed  decimally,  or  in 
the  form  of  a  common  fraction,  and  the  quotient  will  he  the 
cost. 

EXAMPLES    FOR    PRACTICE. 

2.  By  selling  sugar  at  8  cents  a  pound,  a  merchant  lost  20 
per  cent.  ;  what  did  the  sugar  cost  him  ?        Ans.    10  cents. 

3.  Sold  flour  for  §6.12^  per  barrel,  and  lost  12J-  per  cent. ; 
what  was  the  cost  ?  Ans.    $7.00. 

4.  A  grocer,  by  selling  tea  at  $.96  a  pound,  gains  28  per 
cent.  ;  how  much  did  it  cost  him  ?  Ans.    '^.1  b. 

5.  Sold  a  quantity  of  flour  for  $1881,  which  was  ISf  per 
cent,  more  than  it  cost ;  how  much  did  it  cost  ? 

6.  Sold  25  barrels  of  apples  for  $G9.75,  and  made  24  per 
cent. ;  how  much  did  they  cost  per  barrel  ? 

7.  Sold  9i  cwt.  of  sugar  at  $8^  per  cwt.,  and  thereby  lost 
12  per  cent.  ;  how  much  was  the  whole  cost? 

8.  Having  used  a  carriage  six  months,  I  sold  it  for  $9G, 
which  was  20  per  cent,  below  cost ;  what  would  I  have  received 
had  I  sold  it  for  lo  per  cent,  above  cost?  Ans.    $138. 

9.  B  sells  a  pair  of  horses  to  C,  and  gains  12^  per  cent. ; 
C  sells  them  to  D  for  $570,  and  by  so  doing  gains  18J  per 
cent.;  how  much  did  the  horses  cost  B?      Ans.    $42G.66j. 

10.  A  grocer  sold  4  barrels  of  sugar  for  $24  each  ;  on  2 
barrels  he  gained  20  per  cent.,  and  on  the  other  2  he  lost  20 
per  cent. ;  did  he  gain  or  lose  on  the  whole  ?     Ans.   Lost  $4. 

11.  A  person  sold  out  his  interest  in  business  for  $4900, 
which  was  40  per  cent,  more  than  3  times  as  much  as  he  began 
with;  how  much  did  he  begin  with  ?  Ans.    $1166.66|. 

Give  explanation.    Rule. 


INSURANCE.  223 


INSURANCE. 

QSl.  Insurance  on  property  is  security  guaranteed  by 
one  pcirty  to  another,  for  a  stipulated  sum,  against  the  loss  of 
that  property  by  fire,  navigation,  or  any  other  casualty. 

1365.  The  Insurer  or  ITnclerwriter  is  the  party  taking  the 
risk. 

S66.    Tiie  Insured  is  the  party  protected. 

ISIIT.  The  Policy  is  the  written  contract  between  the 
parties. 

S^§.  The  Premium  is  the  sum  paid  by  the  insured  to  the 
insurer,  and  is  estimated  at  a  certain  rate  per  cent,  of  the 
amount  insured,  which  rate  varies  according  to  the  degree  of 
hazard,  or  class  of  risk. 

Note.  As  a  security  against  fraiid,  most  msiirance  companies  take 
risks  at  not  more  than  two  tlni'ds  tlie   full  value  of  the  property 

insured. 

S6^.  To  find  the  premium  when  the  rate  of  insur- 
ance and  the  amount  insured  are  given. 

1.  What  must  I  pay  rinnually  for  insuring  my  house  to  the 
amount  of  $.3250,  at  1^  per  cent,  premium  ? 

opERATiox.  Analysis.      We 

$3250  X  -01^  or  .0125  =  $40.G25.  multiply  the  amount 

Or   11  npr  pt  —  ^&^  — J--  insured,    $3250,   by 

V-^li  ^4   PCI   »-t-  iTSTS  —  so  '  ,  til 

S3250  X  ,V  =  SJ0.C2i.  tltXii^.  'JZ 

$40,625,  is  the  premium.     Or,  the  rate,  11  per  cent.,  is  ^r=  J^  of 
the  amount  insured,  and  -gL  of  $3250  is  $40,621.     Hence  the 

Rule.  3faUiply  the  amount  insured  hj  the  rate  per  cent., 
and  the  product  will  be  the  premium.     Or, 

Take  such  a  part  of  the  amount  insured  as  the  rate  is  part 
of  100. 

Define  insurance.  Insurer,  or  underwriter.  Policy.  Premium. 
To  what  amount  can  property  usually  be  msured  ?  Give  analysis  of 
example  1.     Rule. 


224  PERCENTAGE. 

EXAMPLES   FOR   PRACTICE. 

2.  TVhat  is  the  premium  on  a  policy  for  $750,  at  4  per 
cent.  ?  Ans.    $30. 

3.  "Wliat  premium  must  be  paid  for  $4572.80  insurance,  at 
2J- percent.?  Ans.    $114.32. 

4.  A  house  and  furniture,  valued  at  $5700,  are  insured  at 
If  per  cent. ;  what  is  the  premium  ?  Ans.    $99.75. 

5.  A  vessel  and  cargo,  valued  at  $28400,  are  insured  at  S^ 
per  cent. ;  what  is  the  premium  ?  Ans.    $994. 

6.  A  woolen  factory  and  contents,  valued  at  $55800,  are 
insured  at  2f  per  cent. ;  if  destroyed  by  fire,  what  would  be 
the  actual  loss  of  the  company  ?  Ans.    $54237.60. 

7.  What  must  be  paid  to  insure  a  steamboat  and  cargo 
from  Pittsburg  to  New  Orleans,  valued  at  $47500,  at  f  of  1 
percent.?  Ans.    $356.25. 

8.  A  gentleman  has  a  house,  insured  for  $8000,  and  the  fur- 
niture for  $4000,  at  2 1  per  cent,;  what  premium  must  he 
pay?  Ans.  $285. 

9.  A  cargo  of  4000  bushels  of  wheat,  worth  $1.20  a  bushel, 
is  insured  at  f  of  1^  per  cent,  on  %  of  its  value ;  if  the  cargo  be 
lost,  how  much  will  the  owner  of  the  wheat  lose  ?    Avs.   $1036. 

10.  What  will  it  cost  to  insure  a  factory  valued  at  $21000, 
at  |-  per  cent. ;  and  the  machinery  valued  at  $15400,  at  |  per 
cent.?  A71S.   $264.25. 

TAXES. 

t 
270,     A  Tax  is  a  sum  of  money  assessed  on  the  person 

or  property  of  an  individual,  for  public  purposes. 

371.  When  a  tax  is  assessed  on  property,  it  is  apportioned 
at  a  certain  per  cent,  on  the  estimated  value. 

When  assessed  on  the  person,  it  is  apportioned  equally 
among  the  male  citizens  liable  to  assessment,  and  is  called  a 
poll  tax.     Each  person  so  assessed  is  called  a  poll. 

"S\niat  is  a  tax  ?  How  is  a  ta.\  on  property  apportioned  ?  On  the 
person,  how  ? 


TAXES.  225 

*I73.  Property  is  of  two  kinds  —  real  estate,  and  personal 
property. 

^?»?.  Eeal  Estate  consists  of  immovable  property,  such 
a-  lands,  houses,  itc. 

^74.  Eersonal  Property  consists  of  movable  property, 
such  as  money,  notes,  furniture,  cattle,  tools,  &c. 

^7*5.  An  Inventory  is  a  written  list  of  articles  of  proper- 
ty, with  their  value. 

SI^H.  Before  taxes  are  assessed,  a  complete  inventory  of  all 
the  taxable  property  upon  which  the  tax  is  to  be  levied  must 
be  made.  If  the  assessment  include  a  poll  tax,  then  a  complete 
list  of  taxable  polls  must  also  be  made  out. 

I.  A  tax  of  $3165  is  to  be  assessed  on  a  certain  town; 
the  valuation  of  the  taxable  property,  as  shown  by  the  as- 
sessment roll,  is  $000,000,  and  there  are  220  polls  to  be  as- 
sessed 75  cents  each ;  what  will  be  the  tax  on  a  doUai-,  and 
how  much  will  be  A's  tax,  whose  property  is  valued  at  $3750, 
and  who  pays  for  3  polls? 

OPERATION. 

$.75  X  220  izr  $165,  amount  assessed  on  the  polls. 

$31(35  — $165  =  $3000,  amount  to  be  assessed  on  the  property. 

$3000  -:-  $600,000  =  .005,  tax  on  $1. 

$3750  X  -005  z=.  $18.75,  A's  tax  on  property. 

$.75  X  3  =:  $2.25,  A's  tax  on  3  polls. 

$18.75  +  $2.25  z=  $21,  amount  of  A's  tax. 

Hence  the  following 

Rule.  I.  Find  the  amount  of  poll  tax,  if  any,  and  subtract 
this  sum  from  the  whole  amount  of  tax  to  be  assessed. 

II.  Divide  the  sum  to  be  raised  on  property,  hy  the  xohole 
amount  of  taxable  property,  and  the  quotient  will  be  the  per 
cent.,  or  the  tax  on  one  dollar. 

III.  Midtiply  each  man's  taxable  property  by  the  per  cent., 
or  the  tax  on  $1,  and  to  the  product  add  Jus  poll  tax,  if  any  ; 
the  result  will  be  the  whole  amount  of  his  tax. 

A\Tiat  is  real  estate  ?    Personal  property  ?    An  inventory  ?     Explain 
the  process  of  levying  a  state  or  other  tax.     Rule. 
10* 


22G 


PERCENTAGE. 


Note.  Having  fomicl  the  tax  on  ^'1,  or  the  per  cent.,  -which  in  the 
preceding  example  -vve  find  to  be  5  mills,  or  1  per  cent.,  the  operation 
of  assessing  taxes  may  be  greatly  facilitated  by  finding  the  tax  on  i^2, 
$3,  Sec,  to  ,$10,  and 'then  on  $20,  $30,  &c.,  to  $100,  and  arranguig 
the  nuiiibcrs  as  in  the  following 


TABLE. 


Trop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1  gives 

$.005 

$10 

$.05 

$100 

$  .50 

$1000 

$5.00 

2    " 

.01 

20 

.10 

200 

1.00 

2000 

10. 

3    " 

.015 

30 

.15 

300 

1.50 

3000 

15. 

4    " 

.02 

40 

.20 

400 

2.00 

4000 

20. 

5    " 

.025 

50 

.25 

500 

2.50 

5000 

25. 

6    " 

.03 

60 

.30 

600 

3.00 

6000 

30. 

7    " 

.035 

70 

.35 

700 

3.50 

7000 

35. 

8    " 

.04 

80 

.40 

800 

4.00 

8000 

40. 

9    " 

.045 

90 

.45 

900 

4.50 

9000 

45. 

(( 

(( 

11 

(( 

« 

i( 

(( 

(i 

(( 

(( 

« 

i( 

$15.00 

4.00 

.20 

.025 

1.50 


EXAMPLES    FOR    PRACTICE. 

2.  According  to  the  conditions  of  the  last  example,  liow 
much  would  be  a  person's  tax  whose  property  Avas  assessed 
at  $3845,  and  who  paid  for  2  polls  ? 

Finding  the  amount  from  the  table, 

The  tax  on  $3000 is  .  .  . 

800 " 

40 " 

5 " 

2  polls  " 

Total  tax  is $20,725 

3.  How  mucli  would  be  Ws  tax,  who  was  assessed  for  1 
poll,  and  on  property  valued  at  $5390  ?  A}is.    $27.70. 

4.  A  tax  of  $9190.50  is  to  be  assessed  on  a  certain  village ; 
the  property  is  valued  at  $1400000,  and  there  are  2981  poll:^, 
to  be  taxed  50  cents  each  ;  what  is  the  assessment  on  a  dollar  ? 
what  is  C's  tax,  his  property  being  assessed  at  $12450,  and  ho 
paying  for  2  polls  ?     Ans.  $.005^  on  $1 ;   $09.47^,  C's  tax. 

5.  "What  is  the  tax  of  a  non-resident,  having  property  in 
the  same  village  valued  at  $5375  ?  Ans.    $29.5025. 


Explain  tho  table  and  its  use. 


CUSTOM   HOUSE  BUSINESS.  227 

G.   A  mining   corporation,    consisting   of  30   persons,  are 

taxed  $4342.75;  their  property  is  assessed  for  $188000,  and 

"  each  poll  is  assessed  G2J-  cents ;  what  per  cent,  is  their  tax, 

and  hovv  much  must  he  pay  whose  share  is  assessed  for  $2500, 

and  who  pays  for  1  poll  ?  Ans.  2^^  per  cent.;  $58,125. 

7.  In  a  certain  county,  containing  25482  taxable  inhab- 
itants, a  tax  of  $103294.60  is  assessed  for  town,  county,  and 
state  purposes ;  a  part  of  this  sum  is  raised  by  a  tax  of  30 
cents  on  each  poll ;  the  entire  valuation  of  property  on  the  as- 
sessment roll  is  $38260000  ;  what  per  cent,  is  the  tax,  and  how 
much  will  a  person's  tax  be  who  pays  for  3  polls,  and  whose 
property  is  valued  at  $9470  ?  Ans.  to  last,  $24,575. 

8.  The  number  of  polls  in  a  certain  school  district  is  225, 
and  the  taxable  property  $1246093.75  ;  it  is  proposed  to  build 
a  union  school  house  at  an  expense  of  $10000 ;  if  the  poll  tax 
be  $1.25  a  poll,  and  the  cost  of  collecting  be  2^-  per  cent.,  what 
will  be  the  tax  on  a  dollar,  and  how  much  will  be  E's  tax,  who 
pays  for  1  poll,  and  has  property  to  the  amount  of  $11500  ? 

Ans.    $.008,  tax  on  $1 ;  $93.25,  E's  tax. 

9.  In  a  certain  district  the  school  was  supported  by  a  rate- 
bill  ;  the  teacher's  wages  amounted  to  $200,  the  fuel  and  other 
expenses  to  $75.57 ;  the  public  money  received  was  $98,  and 
the  whole  number  of  -days'  attendance  was  3946  ;  A  sent  2 
pupils  118  days  each;  how  much  was  his  rate-bill  ?  Ans.  $10.62 

CUSTOM  HOUSE  BUSINESS. 

*|'J"7'.  Duties,  or  Customs,  are  taxes  levied  on  imported 
goods,  for  the  support  of  government  and  the  protection  of  home 
industry. 

S78.  A  Custom  House  is  an  office  established  by  govern- 
ment for  the  transaction  of  business  relating  to  duties. 

^7I>.  A  Port  of  Entry  is  a  seaport  town  having  a  cus- 
tom house. 


DeSue  duties.     A  custom  house. 


228  PERCENTAGE. 

980.  Tonnage  is  a  tax  levied  upon  a  vessel,  independent 
of  its  cargo,  for  the  privilege  of  coming  into  a  port  of  entry. 

281.  Revenue  is  the  income  to  government  from  duties 
and  tonnage. 

Duties  are  of  two  kinds  —  ad  valorem  and  specific. 

982.  Ad  Valorem  Duty  is  a  sum  computed  on  the  cost 
of  goods  in  the  country  from  which  they  were  imported. 

983.  Specific  Duty  is  a  sum  computed  on  the  weight  or 
measure  of  goods,  without  regard  to  their  cost. 

284.  An  Invoice  is  a  bill  of  goods  imported,  showing 
the  quantity  and  price  of  each  kind. 

285.  By  the  New  Tariflf  Act,  approved  March  2,  1857, 
all  duties  taken  at  the  U.  S.  custom  houses,  are  ad  valorem. 

In  collecting  customs,  it  is  the  design  of  government  to  tax 
only  so  much  of  the  merchandise  as  will  be  available  to  the 
importer  in  the  market.  The  goods  are  weighed,  measured, 
gauged,  or  imported,  in  order  to  ascertain  the  actual  quantity 
and  value  received  in  port;  and  an  allowance  is  made  in  every 
case  of  waste,  loss,  or  damage. 

286.  Tare  is  an  allowance  of  the  weight  of  the  package 
or  covering  that  contains  the  goods.  It  is  ascertained  by 
actually  weighing  one  or  more  of  the  empty  boxes,  casks,  or 
coverings.  In  common  articles  of  importation,  it  is  sometimes 
computed  at  a  certain  per  cent,  previously  ascertained  by 
frequent  trials. 

287'.  Leakage  is  an  allowance  on  liquors  imported  in 
casks  or  barrels. 

288.  Breakage  is  an  allowance  en  liquors  imported  in 

bottles. 

Note.     Actual  leakage  or  breakage  is  allowed,  tlicre  being  no  fixed 
or  legal  rato. 

289.  Gross  Weight  or  Value  ia  the  weight  or  value  of 
the  goods  before  any  allowance  has  been  made. 

290.  Net  Weight  or  Value  is  the  weight  or  value  after 
all  allowances  have  been  deducted. 


Define  Tonnage.  Revenue.  Ad  valorem  duty.  Spiecifio  duty.  An 
invoice^  Tare.  Leakage.  Breakage.  Gross  weight  or  value.  Net 
■weight  or  value. 


I 


CUSTOM  HOUSE  BUSINESS.  229 

Note.  Draft  is  an  allowance  for  the  waste  of  certain  ai'ticles,  and 
is  made  only  for  statistical  purposes ;  it  does  not  atfect  the  amount  of 
duty.     The  rates  of  this  allowance  are  as  follows : 

On        112  1b , lib. 

Above  112  lb.  and  not  exceeding  224  lb.,  2  lb. 

"      224  lb.     "     "  "       336  lb.,  3  lb. 

"      336  1b.     "     "  "      1120  lb.,  4  1b. 

"    1120  1b.     "     "  "      2016  1b.,  7  lb. 

"    2016  1b 9  1b. 

1.  What  is  the  duty,  at  24  per  cent.,  on  50  gross  of  London 
ale,  invoiced  at  $1.20  per  dozen,  2J  per  cent,  being  allowed 
for  breakage  ? 

OPERATION'.  Analysis.      TVe 

S1.20  X  12  X  50  =  ST20,  gross  value,      first  find  the  cost  of 
$720  X  .025  =  818,  leakage.  the  ale,  at  the  in- 

§720  —  818  =  $702,  net  value.  voice  price,which  is 

$702  X     .24  =  $168.48,  duty.  ^^^O.      From    this 

sum  we  deduct  the 
allowance  for  breakage,  $18,  and  compute  the  duty  on  the  re- 
mainder.    Kence  the  following  ^ 

KuLE..  Deduct  allowances,  if  necessary,  and  compute  the 
duty,  at  tlie  given  rate,  on  the  net  value. 

Note. — In  the  following  examples,  the  legal  rate  of  duty  will  be 
given,  according  to  the  Tariff  of  1851. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  duty  at  19  per  cent,  on  224  yards  of  plaid 
silk,  invoiced  at  $.95  per  yard  ?  Ans.  $40.43  + . 

8.  What  is  the  duty  at  24  per  cent,  on  50  barrels  of  sperm 
oil,  each  containing  originally  31 J  gallons,  invoiced  at  $.54  per 
gallon,  allowing  2  per  cent,  for  leakage?       Ans.  $200.03  +  . 

4.  What  is  the  duty  at  15  per  cent,  on  175  bags  of  Java 
coffee,  each  containing  115  lbs.,  valued  at  15  cents  per  pound? 

Ans.  1452.81+ 

5.  John  Jones  imported  from  Havana  25  hhds.  of  W.  I. 
molasses,  which  was  invoiced  at  36  cents  per  gallon ;  allowing 
5  per  cent,  for  leakage,  what  was  the  duty  at  24  per  cent.  ? 

Ans.  $135,399  +  . 

Define  draft.     Give  analysis.     Rule. 


230  SIMPLE   INTEREST. 

SIMPLE  INTEREST. 

S91«    Interest  is  a  sum  paid  for  the  use  of  money. 

*^i>S.  Principal  is  the  sum  for  the  use  of  which  interest 
is  ])aid. 

!!l9el.  Eate  per  cent,  per  annum  is  the  sum  per  cent,  paid 
for  t!ie  use  of  $100  annually. 

Note.     The  rate  per  cent,  is  commonly  expressed  decimally,  as  him- 
dredths  (331). 

!S!>4.    Amoiint  is  the  sum  of  the  principal  and  interest. 

^9<5.  Simple  Interest  is  the  sum  paid  for  the  use  of  the 
principal  only,  during  the  whole  time  of  the  loan  or  credit. 

S®©.  Legal  Interest  is  the  rate  per  cent,  established  by 
law.     It  varies  in  different  States,  as  follows : 


Alabama, 8  per  cent. 

Arkansas, 6 

Connecticut, G 

Delaware, 6 

Dist.  of  Columbia,  ...  6 

Florida, 8 

Georgia, 7 

Illinois, 6 

Indiana, 6 

Iowa, 7 

Kentucky, 6 

Louisiana 5 

Maine, (3 

jNIaryland, 6 

Massachusetts, 6 

Michigan, 7 


(( 

(( 

li 

l( 

1( 

(( 

(( 

(( 

(( 

l< 

l( 

(1 

(i 

(t 

« 

Mississippi, 8  per  cent. 

Missouri, G     "      " 

New  Hampshire,  . . .  .G     "      " 

New  Jersey, 6     "      " 

New  York, 7     "      " 

North  Carolina, 6     "      " 

Ohio, G     "      " 

Pennsylvania, G     "      " 

lihode  Island, G     "      " 

South  Carolina, 7 

Tennessee, G 

Texas, 8 

United  States  (debts),  6 

Vermont, 6 

Virginia, 6 

Wisconsin, 7 


Note,     ^^^len  the  rate  per  cent,  is  not  spccilicd,  in  accoimts,  notes, 
mortgages,  contracts,  &c.,  the  legal  rate  is  always  understood. 

29t,   Usury  is  illegal  interest,  or  a  greater  jier  cent,  tha.i 
the  legal  rate. 

CASE   I. 

998.    To  find  the  interest  on  any  sum,  at  any  rate 
per  cent.,  fur  years  and  months. 

Define  interest.     Principal.     Rate  per  cent,  per  annum.     Amount. 
What  is  simple  interest  ?     Legal  interest  ?     Usury  ?     Case  I  ? 


PERCENTAGE.  231 

In  i^Grcentage,  anj'  per  cent,  of  any  given  number  is  so 
many  hundredtlis  of  that  number  ;  but  in  interest,  any  rate  per 
cent,  is  confined  to  1  year,  and  the  per  cent,  to  be  obtained 
of  any  given  number  is  greater  tlian  the  rate  per  cent  per 
annum  if  the  time  be  7nore  tlian  1  year,  and  less  than  the  rale 
per  cent,  per  annum  if  the  time  be  less  tlian  1  year.  Thus, 
tlie  interest  on  any  sum,  at  any  rate  per  cent.,  for  3  yeai-s  G 
montlis,  is  3i  times  the  interest  on  the  same  sum  for  1  year; 
and  tlie  interest  for  3  months  is  ^  of  the  interest  for  1  year. 

1.  Wliat  is  the  interest  on  S75.19  for  3  years  G  months,  at 
G  per  cent.  ? 

OPERATIOX. 

$7o.l9 

.06  Analysis.    The  interest  on  $75.19,  for  1  yr., 

^  at  6  per  cent.,  is  .06  of  the  principal,  or  $4.5114, 

'  and  the  interest  for  3  yr.  6  mo.  is  3  A  ^r  3^  times 

^2  the  interest  for  1  jr.,  or  $4.5114  X  3-^,  which  is 


2-557  $15,789  -j-,  the  Ans.     Hence,  the  following 

135342 


$15.7899,  Ans. 

Rule.  I.  Multiply  the  principal  hy  the  rate  per  cent,  and 
the  product  loill  be  the  interest  for  1  year. 

II.  Multiply  this  product  hy  the  time  in  years  and  fractions 
of  a  year,  and  the  result  will  be  the  required  interest. 

EXAMPLES    FOR   PPvACTICE. 

2.  What  is  the  interest  of  $150  for  3  years,  at  4  per  cent.  ? 

Ans.    $18. 

3.  What  is  the  interest  of  $328  for  2  years,  at  7  per  cent.  ? 

4.  What  is  the  interest  of  $125  for  1  year  G  months,  at  6 
percent.?  Ans.    $11.25. 

5.  What  is  the  interest  of  $200  for  3  years  10  months,  at 
7  per  cent  ?  Ans.    $53. G6  -f- . 

6.  What  is  the  interest  of  $76.50  for  2  years  2  months,  at 

5  per  cent.  ?  Ans. .  $8,287  -f  . 

Explain  the  difference  between  percentage  and  interest.  Give 
analysis.     Rule. 


232  SIMPLE   INTEREST. 

7.  What  is  the  interest  of  $1276.25  for  11  months,  at  7 
percent.?  Ans.    $81.89  +  . 

8.  What  is  the  interest  of  $2569.75  for  4  years  6  months, 
at  6  per  cent.  ? 

9.  What  is  the  interest  of  $1500.60  for  2  years  4  months, 
at  6^  per  cent.  ?  Ans.   $218.8375. 

10.  What  is  the  amount  of  $26.84  for  2  years  6  month?,  at 
5  percent.?  Ans.    $30,195. 

11.  What  is  the  amount  of  $450  for  5  years,  at  7  per  cent.  ? 

12.  What  is  the  interest  of  $4562.09  for  3  years  3  months, 
at  3  per  cent.  ?  A7is.    $444.80  +  . 

13.  What  is  the  amount  of  $3050  for  4  years  8  months,  at 
5^  per  cent.  ?  Ans.  $3797.25  +  . 

14.  What  is  the  interest  of  $5000  for  9  months,  at  8  per 
cent.?  Ans.    $300. 

15.  If  a  person  borrow  $375  at  7  per  cent.,  liow  mucli  will 
be  due  the  lender  at  the  end  of  2  yr.  6  mo.  ? 

16.  What  is  the  interest  paid  on  a  loan  of  $1374.74,  at  6 
per  cent.,  made  January  1,  1856,  and  called  in  January  1, 
18G0?  Ans.    $329,937  +  . 

17.  If  a  note  of  $605.70  given  May  20, 1858,  on  interest  at 
8  percent.,  be  taken  up  May  20,  1861,  what  amount  will  then 
be  due  if  no  interest  has  been  paid?     Ans.   $751,068. 

CASE    II. 

299.  To  find  the  interest  on  any  sum,  for  any 
time,  at  any  rate  per  cent. 

The  analysis  of  our  rule  is  based  upon  the  following 

Obvioics  Relations  between  Time  mid  Interest. 

I.  The  interest  on  any  sum,  for  1  year,  at  1  per  cent.,  is 
.01  of  that  sum,  and  is  equal  to  the  principal  with  the  scparatrix 
removed  two  places  to  the  left. 

II.  A  month  being  j'j  of  a  year,  yV  of  the  interest  on  any 
sum  for  1  year  is  the  interest  for  1  month. 

"What  is  Case  II  ?  Give  the  first  relation  between  time  and  interest. 
Second. 


PERCENTAGE.  233 

III.  The  interest  on  any  sum  for  3  days  is  t^^j  =z  -^\j  =  .1 
of  the  interest  for  1  month,  and  any  number  of  days  may 
readily  be  reduced  to  tetiihs  of  a  month  by  dividing  by  3. 

IV.  The  interest  on  any  sum,  for  1  month,  multiplied  by 
any  "■iven  time  expressed  in  months  and  tenths  of  a  month, 
will  produce  the  required  interest. 

I.  What  is  the  interest  on  S724.G8  for  2  yr.  5  mo.  19  da., 
at  7  per  cent.  ? 

OPERATION.  Analysis.      We  remove 

2yr.  5mo.  19da.  =  29.G^mo.       the  separatrix  in  the  given 

in  \  (S'Toi/'Q  principal  two  places  to  the 

^ leit,  and  we   have   §/.24(j8, 

$.6039  the  interest  on  the  given  sum 

29.64-  for    1    year   at    1    per   cent. 

-— —  (300   I.).     Dividing  this  by 

"^^^'^  12,  we  have  $.6039,  the  inter- 

36234:  est  for  1  month,  at  1  per  cent. 

54351  (II.)       Multiplying    this 

12078  quotient  by  '29.6^,  the  time 

'  ~  expressed  in  months  and  deci- 

$17.895o7  jj^j^jg  Qf  ^  j^Qj^^i^^  (jjj_  jy_^) 

we  have  $17.89557,  the  in- 

$125.26899,  Ans.  terest  on  the  given  sum  for 

the  given  time,  at  1  per  cent. 
(rV.).  And  multiplying  this  product  by  7  (7  times  1  per  cent.), 
we  have  $125,208  -|-,  the  interest  on  the  given  principal,  for  the 
given  time,  at  the  given  rate  per  cent.     Hence, 

Rule.  I.  Remove  the  separatrix  in  the  given  principal 
tivo  places  to  the  left  ;  the  result  will  he  the  interest  for  1  year, 
at  1  per  cent. 

II.  Divide  this  interest  hy  12  ;  the  result  ivill  be  the  interest 
for  1  months  at  1  per  cent. 

III.  Multiply  this  interest  hy  the  given  time  expressed  in 
months  and  tenths  of  a  month  ;  the  result  will  he  the  interest 
for  the  given  time,  at  1  per  cent. 

IV.  Multiply  this  interest  hy  the  given  rate  ;  the  product 
will  be  the  interest  required. 

Give  the  thh-d.     Fomth.     Give  analysis.     Rule. 


234  SIMPLE  INTEREST. 

Contractions.  After  removing  the'  separatrix  in  the  principal 
two  places  to  the  left,  the  result  may  be  regarded  either  as  the  in- 
terest on  the  given  principal  for  12  months  at  1  per  cent.,  or  for  1 
month  at  12  per  cent.  If  we  regard  it  as  for  1  mouth  at  12  per 
cent.,  and  if  the  given  rate  be  an  aliquot  part  of  12  per  cent.,  the 
interest  on  the  given  principal  for  1  month  may  readily  be  found  by 
taking  such  an  aliquot  part  of  the  interest  for  1  month  as  the  given 
rate  is  part  of  12  per  cent.     Thus, 

To  find  the  interest  for  1  month  at  G  per  cent.,  remove  the  sep- 
aratrix two  places  to  the  left,  and  divide  by  2. 

To  find  it  at  3  per  cent.,  proceed  as  before,  and  divide  by  4  ;  at  4 
per  cent.,  divide  by  3  ;  at  2  per  cent.,  divide  by  6,  &c. 

SIX    PER   CENT.    METHOD. 

300.  By  referring  to  S9@  it  will  be  seen  that  the  legal 
rate  of  interest  in  21  States  is  6  per  cent.  This  is  a  sufficient 
reason  for  introducing  the  following  brief  method  into  this  work  : 

Analysis.     At  6  per  cent-  per  annum  the  interest  on  $1 

For  12  months is  $.06. 

"      2  months  {j%-=i-  of  12  mo.) "    .01. 

"      1  month,  or  30  days( yL  of  12  mo.)  "    .00J=$.005  {^^  of  $-0G)- 

"      6  days  (I  of  30  days)" "    .001. 

"      1     "     (>-of  Gda.==gVof30days)  "    .000^. 
Hence  we  conclude  that, 

1st.  The  interest  on  $1  is  S.005  per  mouth,  or  ^.01  for  every 
2  months ; 

2d.  The  interest  on  $1  is  ^.OOOi  per  day,  or  8.C01  for  every 
G  days. 

From  these  principles  we  deduce  the 

Rule.  I.  To  find  the  rate  :  —  Call  every  year  $.0G,  every 
2  months  $.01,  every  6  days  $.001,  and  any  less  number  o/  days 
sixtJis  of  1  mill. 

II.  To  find  the  interest : — Jltdttply  the  2)rincqKd  hy  the  rate. 

Notes. — 1.  To  find  the  interest  at  any  other  rate  per  cent,  b^-  this 
method,  first  find  it  at  G  per  cent.,  and  tlien  increase  or  diminish  the 
result  by  as  many  tnnes  itself  as  the  given  rate  is  grcator  or  less  than  6 
per  cent.     Thus,  for  7  per  cent,  add  i^,  for  4  per  cent,  subtract  ^,  &c. 

What  contractions  are  given?  Giveanalj'sisof  the  C  per  cent,  method. 
Rule.     Its  application  to  any  other  rate  ])er  cent. 


SIMPLE  INTEREST.  235 

2.  The  interest  of  $10  for  6  clays,  or  of  $1  for  GO  ilays,  is  $.01. 
Therefore,  if  the  principal  be  less  thari  $10  ami  the  tiino  less  than  (i 
days,  or  the  principal  less  than  $1  and  tlie  time  less  than  GO  days,  the 
interest  will  be  less  tlian  $.01,  and  may  be  disregardeu. 

3.  Since  the  interest  of  $1  for  GO  days  is  $.01,  the  interest  of  $1  for 
any  number  of  days  is  as  many  cents  as  60  is  contained  times  in  the 
number  of  days.  Therefore,  if  any  principal  be  multiplied  by  tlie  num- 
ber of  days  in  any  given  number  of  moutlis  and  days,  and  tlie  proiluct 
divided  by  GO,  the  result  will  be  the  interest  in  cents.  That  is,  Jfii'ii- 
j^ly  the  principal  by'  the  number  of  days,  divide  the  product  bi/  GO,  and  punit 
ojf  two  decimal  places  in  the  quotient.  The  result  will  be  the  interest  in  ihe 
same  daiuwination  as  th&  principal. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  interest  of  $100  for  7  years  7  months,  at  6 
per  cent.  ?  Ans.    $45.50. 

3.  What  is  the  amount  of  $47.50  for  4  years  1  month,  at  9 
per  cent.  ?  j47is.    $G4.95G  -j-  . 

4.  What  is  the  amount  of  $2000  for  3  months,  at  7  per 
cent.?  ^  '  Ans.    $2035. 

5.  AVhat  is  the  interest  of  $250  for  1  year  10  months  and 
15  days,  at  G  per  cent.  ?  A7is.   $28.12^. 

G,    What  is  the  interest  of  $30.75  for  2  years  4  months  and 
12  days,  at  7  per  cent.  ?  Ans.    $G.088  +  . 

7.  What  is  the  amount  of  $84  for  5  years  5  months  and  9 
day.«!,  at  5  per  cent.  ? 

8.  What  is  the  interest  of  $51.10   for  10  months  and  3 
days,  at  4  per  cent.  ? 

9.  What  is  the  interest  of  $175.40  for  15  morrths  and  8 
days,  at  10  per  cent.  ?  Ans.    $22.31  +  . 

10.  What  is  the  amount  of  $1500  for  G  months  and  24 
days,  at  7^  per  cent.  ?  Ans.    $1563.75. 

11.  What  is  the  amount  of  $84.25  for  1  year  5  months 
and  10  days,  at  6|-  per  cent.  ? 

12.  What  is  the  interest  of  $25  for  3  years   6  months  and 
20  days,  at  G  per  cent.  ?  Ans.    $5.33^. 

13.  What  is  the  interest  of  $112.50  for  3  months  and  1 
4ay,  at  9^  per  cent.  ?  Ans.    $2.70  -}-• 

What  contractions  are  given  ? 


236  PERCENTAGE. 

14.  What  is  the  interest  of  $408  for  20  days,  at  G  per 
cent.?  Ans.    $1.36. 

15.  "What  is  the  interest  of  $500  for  22  days,  at  7  per 
cent.  ? 

IG.  What  is  the  amount  of  $4500  for  10  days,  at  10  per 
cent.?  Ans.    $4512.50. 

17.  What  is  the  amount  of  $1000  for  1  month  5  days,  at 
Cf])ercent.  ?  A>is.    $100G.56i. 

18.  Find  the  interest  of  $973.G8  for  7  months  9  days,  at 
4-1  per  cent. 

19.  If  I  borrow  $275  at  7  per  cent.,  how  much  will  I  owe 
at  the  end  of  4  months  25  days  ? 

20.  A  person  bought  a  piece  of  property  for  $2870,  and 
agreed  to  pay  for  it  in  1  year  and  G  months,  with  GJ-  per  cent, 
interest;  what  amount  did  he  pay  ?  A)7s.    $3149.825. 

21.  In  settling  with  a  merchant,  I  gave  my  note  for  $97.75, 
due  in  11  months,  at  5  per  cent.;  what  must  be  paid  when 
the  note  falls  due  ?  A)is.    $102.23+. 

22.  How  much  is  the  interest  on  a  note  of  $384.50  in  2 
years  8  months  and  4  days,  at  8  per  cent.  ? 

23.  What  is  the  interest  of  $97.86  from  May  17,  1850,  to 
December  19,  1857,  at  7  per  cent.  ?  Ans.    $51.98  -f-. 

24.  Find  the  interest  of  $35.G1,  from  Nov.  11,  1857,  to 
Dec.  15,  1859,  at  G  per  cent.  Ans.    $4,474. 

25.  Required  the  interest  of  $50  from  Sept.  4,  1818,  to 
Jan.  1,  1860,  at  3^  per  cent. 

26.  Required  the  amount  of  $387.20,  from  Jan.  1  to  Oct. 
20,  1859,  at  7  per  cent.  Ans.    $408,957  -f . 

27.  A  man,  owning  a  furnace,  sold  it  for  $6000 ;  the  terms 
were,  $2000  in  cash  on  delivery,  $3000  in  9  months,  and  llie 
remainder  in  1  year  6  months,  witli  7  per  cent,  interest; 
what  was  the  whole  amount  paid  ?  Ans.    $G2G2.50. 

28.  Wm.  Gallup  bought  bills  of  dry  goods  of  Geo.  Bliss 
&  Co.,  of  New  York,  as  follows,  viz.:  Jan.  10,  1858,  $350; 
April  15,  1858,  $150  ;  and  Sept.  20,  1858,  $550.50  ;  he  bought 
on  time,  paying  legal  interest ;  what  Avas  the  whole  amount 
of  his  indebtedness  Jan.  1,  1859?  Ans.    $1092.66  +. 


PARTIAL   PAYMENTS.  237 


PARTIAL   PAYMENTS   OR   INDORSEMENTS. 

S©  3 .  A  Partial  Payment  is  payment  in  part  of  a  note, 
bond,  or  other  obligation  ;  when  the  amount  of  a  payment  is 
written  on  the  back  of  tlie  obligation,  it  becomes  a  receipt,  and 
is  called  an  Indorsement. 

$2000.  Springfield,  Mass.,  Jan.  4,  1857. 

1.  For  value  received  I  promise  to  pay  James  Parish,  or 
order,  two  thousand  dollars,  one  year  after  date,  with  interest. 

George  Jones. 
On  this  note  were  indorsed  the  following  payments : 

Feb.  19,  1858, $400 

June  29,  1859, $1000 

Nov.  14,  1859, $520 

What  remained  due  Dec.  24,  1860? 

OPERATION. 

Principal  on  interest  from  Jan.  4,  1857,   $2000 

Interest  to  Feb.  19,  1858,  1  yr.  1  mo.  15  da., 135   " 

Amount, $2135 

Payment  Feb.  19,  1858, 400 

Remainder  for  a  new  principal, $1735 

Interest  from  Feb.  19,  1858,  to  June  29,   1859,  1  yr. 

4  mo.  10  da., 141.69 

Amount, $LS7G.69 

Payment  June  29,  1859, 1000. 

Eemainder  for  a  new  principal, $876.69 

Interest  from  June  29,  1859,  to  Nov.  14,  1859,  4  mo. 

15  da., 19.725 

Amount, $896,415 

Payment  Nov.  14,  1859, 520. 

Remainder  for  a  new  principal, $376,415 

Interest  from  Nov.  14,  1859,  to  Dec.  24,  1860,  1  yr. 

1  mo.  10  da., 25.09 

Remains  due  Dec.  24,  1860 $401.505 -f- 

What  is  meant  by  partial  payment  ?     By  an  indorsement  ? 


233  PERCENTAGE. 


^-^^^•^Q-  New  York,  May  1,  1855. 

2.  For  value  received,  we  jointly  and  severally  promise  to 
pay  Mason  &  Bro.,  or  order,  four  hundred  seventy-five  dol- 
lars fifty  cents,  nine  months  after  date,  v.'ith  interest. 

Jones,  Smith  &  Co. 

The  following  indorsements  were  made  on  this  note : 

Dec.  25,  1855,  received, $50 

July   10,  185G,        "         15.75 

Sept.    1,  1857,        "         25.50 

June  14,  1858,         "         104 

How  much  was  due  April  15,  1859  ? 

OPEKATION. 

Principal  on  interest  from  ^lay  1,  1855, $475.50 

Interest  to  Dec.  25,  1855,  7  mo.  24  da., 21.63 

Amount, $497.13 

Pajinent  Dec.  25,  1855, 50. 

Ijlemainder  for  a  new  principal, $447.13 

Interest  from  Dec.  25,  1855,  to  June  14,  1858,  2  yr. 

5  mo.  19  da., 77.29 

Amount, $524,42 

PajTnent  July  10,  1856,  less  than  interest^ 

then  due, >  .$15.75 

Pa}Tnent  Sept.  1,  1857, )    25.50 

Their  sum  less  than  interest  then  due, . . .     $41.25 
Paj-ment  June  14, 1858, 104. 

Their  sum  exceeds  the  interest  then  due, $145.25 

Remainder  for  a  new  principal, $379.17 

Interest  from  June  14,  1858,  to  April  15,  1859,  10  mo. 

1  da., 22.19 

Balance  due  April  15,  1859, $401.36  -|- 

These  examples  liave  been  wrouglit  according  to  llic  method 
prescribed  by  the  Supreme  Court  of  the  U.  S.,  and  are  suf' 
ficient  to  illustrate  the  following 


PARTIAL   PAYMENTS.  289 

United  States  Rule. 

I.  Find  the  amount  of  the  given  principal  to  the  time  of  the 
frst  payment,  and  if  this  payment  exceed  the  interest  then  due 
subtract  it  from  the  amount  obtained,  and  treat  the  remainder 
as  a  neiv  principal. 

II.  But  if  the  interest  be  greater  than  any  payment,  cast  the 
interest  on  the  same  principal  to  a  time  when  the  sum  of  the 
payments  shall  equal  or  exceed  the  interest  due  ;  subtracting  the 
sum  of  the  payments  from,  the  amount  of  the  principal,  the  re- 
mainder will  form  a  new  principcd,  on  ivkich  interest  is  to  be 
computed  as  before. 

^^^^•'^^-  San  Francisco,  June  20,  1858. 

3.  Three  years  after  date  we  promise  to  pay  Ross  & 
"Wade,  or  order,  five  hundred  fourteen  and  -^^^  dollars,  for 
value  received,  with  10  per  cert,  interest.     Wilder  &  Bro. 

On  this  note  Avere  indorse  1  the  folloAving  payments  :'  Nov. 
12,  1858,  $105.50  ;  March  20,  18G0,  S200  ;  July  10,  1860, 
$75.60.  How  much  remains  due  on  the  note  at  the  time  of 
its  maturity?  Ans.    $242.12 -f. 

^^^^"^-  Charleston,  May  7,  1859. 

4.  For  value  received,  I  promise  to  pay  George  Babcock 
three  thousand  dollars,  on  ^demand,  with  7  per  cent,  interest. 

John  May. 

On  this  note  were  indorsed  the  following  payments  :  — 

'  Sept.  10,  1859,  received 625 

-Jan.  1,  1860,  "        500 

Oct.  25,  1860,  "       75 

April  4,  1861,  "        1500 

How  much  was  due  Feb.  20, 1862  ?     Ans.    $1344.35  -f . 


Give  the  ITnited    States  Court  rule  for  computing  interest  Adhere 
partial  pajTuents  have  been  made. 


2-iO  PERCENTAGE. 


^QI'^titV  New  Orleans,  Aug.  3,  1850. 

5.  One  year  after  date  I  promise  to  pay  George  Bailey,  or 
order,  nine  hundred  twelve  -^jj  dollars,  with  5  per  cent,  in- 
terest, for  value  received.  James  Powell. 

The  note  was  not  paid  when  due,  but  was  settled  Sept.  15, 
1853,  one  payment  of  §250  having  been  made  Jan.  1,  1852, 
and  another  of  $316.75,  May  4,  1853.  How  much  was  due 
at  the  time  of  settlement  ?  Ans.    $4G7.53  -|-  . 


^l^^-5^^-  Cincinnati,  April  2,  1860. 

6.  Four  months  after  date  I  promise  to  pay  J.  Ernst  & 
Co.  one  hundred  eighty-four  dollars  fifty-six  cents,  for  value 
received.  S.  Anderson. 

The  note  was  settled  Aug.  26,  1862,  one  payment  of  §50 
having  been  made  May  6,  1861.  How  much  was  due,  legal 
interest  being  6  per  cent.  ?  Atis.    §154.188  -\-  . 

Note.  A  note  is  on  interest  after  it  becomes  due,  if  it  contain  no 
mention  of  interest. 

7.  Mr.  B.  gave  a  mortgage  on  liis  farm  for  §6000,  dated 
Oct.  1,  1851,  to  be  paid  in  6  years,  with  8  per  cent,  interest. 
Three  months  from  date  he  paid  $500  ;  Sept.  10,  1852,  $1126  ; 
March  31,  1854,  $2000  ;  and  Aug.  10,  1854,  $876.50.  How 
much  was  due  at  the  expiration  of  the  time  ?  Ans.  $3284.84  -f- . 

S03.  The  United  States  rule  "for  partial  payments  has 
been  adopted  by  nearly  all  the  States  of  the  Union  ;  the  only 
prominent  exceptions  are  Connecticut,  Vermont,  and  New 
Hampshire. 

Connecticut  Rule. 

I.  Payments  made  one  year  or  more  from  the  time  the  in- 
terest commenced,  or  from  another  payment,  and  payments  less 
than  the  interest  due,  are  treated  according  to  the  United  States 
ride. 


Gire  Connecticut  rule  for  partial  pajments. 


PARTIAL   PAYJIENTS.  241 

II.  Payments  exceeding  the  Interest  due,  and  made  within 
one  year  from  the  time  interest  commenced,  or  from  a  former 
payment,  shall  draw  interest  for  the  balance  of  the  year,  jjro- 
vided  the  interval  does  not  extend  beyond  the  settlement,  and  the 
amount  must  be  subtracted  from  the  amount  of  the  principal  for 
one  year ;  the  remainder  will  be  the  new  principal. 

III.  If  the  year  extend  beyond  the  settlement,  then  find  the 
amount  of  the  payment  to  the  day  of  settlement,  and  subtract  it 
from  the  amount  of  the  principal  to  that  day  ;  the  remainder 
10 ill  be  the  sum  due. 


^^^'^-  Woodstock,  Ct.,  Jan.  1,  1858. 

1.  For  value  received,  I  promise  to  pay  Henry  Bowen,  or 
order,  four  hundred  sixty  dollars,  on  demand,  with  interest. 

James  Marshall. 

On  this  note  are  indorsed  the  following  payments  :  April 
16,  1858,  $148  ;  March  11,  1860,  $75  ;  Sept.  21,  1860,  $56. 
How  much  was  due  Dec.  11,  1860?  Ans.    $238.15-}-. 

30«l.  A  note  containing  a  promise  to  pay  interest  an- 
nucdly  is  not  considered  in  law  a  contract  for  any  thing  more 
than  simple  interest  on  the  principal.  For  partial  payments 
on  such  notes,  the  following  is  the 

Vermont  Rule. 

I.  Find  the  amount  of  the  principal  from  the  time  interest 
commenced  to  the  time  of  settlement. 

II.  Find  the  amount  of  each  payment  from  the  time  it  was 
made  to  the  time  of  settlement. 

III.  Subtract  the  sum  of  the  amounts  of  the  payments  from 
the  amount  of  the  principal,  and  the  remainder  will  be  the 
sum  due. 


S600. 


Rutland,  April  11,  1856. 


1.    For  value  received,  I  promise  to  pay  Amos  Cotting,  or 
order,  six  hundred  dollars  on  demand,  with  interest  annually. 

John  Brown. 


Give  the  Connecticut  rule  for  partial  payments.     The  Vermont  rule. 
E.P  11 


242  PERCENTAGE. 

On  this  note  were  indorsed  the  following  payments  :  Aug. 
10,  1856,  $156;  Feb.  12,  1857,  $200;  June  1,  1858,  $185. 
What  was  due  Jan.  1,  1859  ?  Ans.    $105.50-j-. 

S^4.  In  New  Hampshire  interest  is  allowed  on  the  an- 
nual interest  if  not  paid  when  due,  in  the  nature  of  damages 
for  its  detention ;  and  if  payments  are  made  before  one  year's 
interest  has  occurred,  interest  must  be  allowed  on  such  pay- 
ments for  the  balance  of  the  year.     Hence  the  followinrf 

New  Hampshire  Rule. 

I.  Find  the  amount  of  the  principal  for  one  year,  and  de^ 
duct  from  it  the  amount  of  each  payment  of  that  year,  from  the 
time  it  was  made  up  to  the  end  of  the  year  ;  the  remainder  will 
he  a  new  principal,  with  which  proceed  as  before. 

II.  If  the  settlement  occur  less  than  a  year  from  the  last  an- 
nual term  of  interest,  malce  the  last  term  of  interest  a  jiart  of  a 
year,  accordingly. 

^^^^-  Keene,  N.  H.,  Aug.  4,  1858. 

1.  For  value  received,  I  promise  to  pay  George  Cooper,  or 
order,  five  hundred  seventy-five  dollars,  on  demand,  with  in- 
terest annually.  David  Greekman. 

On  this  note  were  indorsed  the  following  payments  :  Nov. 
4,  1858,  $64;  Dec.  13,  1859,  $48;  March  16,  1860,  S248 ; 
Sept.  28,  1860,  $60.  What  was  due  on  the  note  Nov.  4, 
ISfiO?  Ans.    $215.3^. 

3^«5.  When  no  payment  whatever  is  made,  upon  a  note 
promising  annual  interest,  till  the  day  of  settlement,  in  New 
Hampshire  the  following  is  the 

Court  Rule. 
Compute  separately  the  interest  on  the  principal  from  the 
time  the  note  is  given  to  the  time  of  settlement,  and  the  interest 
on  each  year's  interest  from  the  time  it  should  be  paid  to  the 
time  of  settlement.  The  sum  of  the  interests  thus  obtained, 
added  to  the  principal,  will  he  the  sum  due. 

The  New  Ilampshire  rule.    The  New  Hampshire  court  ride. 


PAKTIAL   PAYMENTS.  243 


?:^  ^       Keexe,  N.  H.,  Feb.  2,  1855. 

1.  Three  years  after  date,  I  promise  to  pay  James  Clark, 
or  order,  five  hundred  dollars,  for  value  received,  with  interest 
annually  till  paid.  ■  John  S.  Briggs. 

What  is  due  on  the  above  note,  Aug.  2, 1859  ?  A71S.  $G49.40. 

Problems  in  Interest. 

SI06.  In  examples  of  interest  there  are  five  parts  involved, 
the  Principal,  the  Eate,  the  Time,  the  Interest,  and  thJ 
Amount. 

CASE   I. 

11©?.    The  time,  rate  per  cent.,  and  interest  being 
given,  to  find  the  principal. 

1.  What  principal  in   2  years,  at   6  per  cent.,  will  gain 
$31.80  interest  ? 

OPERATION.  Analysis.     Since  $1,  in 

$.12,  interest  of  $1  in  2years  atGpercent.  2  years,  at  6  per  Cent.,  AviU 

$31.80  -^  .12  =  $265,  Ans.  S"i'^  8.12  interest,  the  prin- 

cipal thatAvill  gain  $31.80, 
at  tlie  same  rate  and  time,  must  be  as  many  dollars  as  $.12  is  con- 
tained times  in  $31.80;  dividing,  we  obtain  $265,  the  required 
principal.     Hence, 

Rule.     Divide  the  given  interest  hy  the  interest  of  $1  for 
the  given  time  and  rate,  and  the  quotient  will  he  the  -principal. 

EXAMPLES    FOR    PRACTICE. 

2.  What  principal,  at  G  per  cent.,  will  gain  $28. 12^-  in  6 
years  3  months.  ?  Ans.    $75. 

3.  What  sum,  put  at  interest  for  4  months  18  days,  at  4 
per  cent.,  will  gain  $9.20  ?  Ans.    $600. 

4.  What  sum  of  money,  invested  at  7  per  cent.,  will  pay 
me  an  annual  income  of  $1260  ?  Ans.    $18000. 

5.  What  sum  must  be  invested  in  real  estate,  yielding  10 
per  cent,  profit  in  rents,  to  produce  an  income  of  $3370  "^ 

Ans.    $33700. 

How  many  parts  are  considered   in  examples  in  interest  f      What 
are  they  ?    What  is  Case  I  ?     Give  analysis.    Rule. 


214  PERCENTAGE. 

CASE    II. 

308.  The  time,  rate  per  cent.,  and  amount  being 
given,  to  find  the  prineipal. 

1.  AVliat  principal  in  2  years  G  months,  at  7  per  cent., 
will  amount  to  $88,125? 

OPEEATiON.  Analysis. 

$1,175  Amt.  of  $1  in  2  years  6  months,  at  7  per  cent.  Since   $1,  ill 

$88,125  H-  1.175  ==  $75,  Ans.  ^    ^'^,'''\  t 

'  montlis,  at  7 

per  cent.,  will  amount  to  81.17.J,  the  principal  that  will  amount  to 

$88,125,  at  the  same  rate  and  time,  must  be  as  many  dollars  as 

$1,175  is  contained  times  in  $88,125  ;  diviihng,  we  obtain  $75,  the 

reqmred  principal.     Hence  the 

Rule.  Divide  the  given  amount  hj  the  amount  of  $1  _/br 
the  given  time  and  rate,  and  the  quotient  will  he  the  jjrincipal 
required.  Wk* 

EXAMPLES    FOR    TRACTICE.  ' 

2.  "What  principal,  at  G  per  cent.,  will  amount  to  $655.20       .,_ 
in  8  months  ?  ^  Ans.    $630.       ^| 

3.  What  principal,  at  5  per  cent.,  will  amount  to  $106,855       jj 
in  5  years  5  months  and  9  days  ?  Ans.    $81. 

4.  "What  sum,  put  at  interest,  at  5J-  per  cent.,  for  8  years 
5  months,  will  amount  to  $1897.545  ?  Ans.    $1297.09+. 

5.  What  sura,  at  7  per  cent.,  will  amount  to  $221,075  in  3     , 
years  4  months  ?  Ans.    $179.25. 

G.  What  is  the  interest  of  that  sum,  for  11  years  8  days,  at 
10^-  per  cent.,  which  will  at  the  given  rate  and  time  amount  to 
$857.54?  Ans.    $460.04. 

CASE   III.  P 

JJ01>.  The  principal,  time,  and  interest  being  given, 
to  find  *Jie  rate  per  cent.  ,|| 

1.  I  lent  $450  for  3  years,  and  received  for  interest  $G7.50; 
what  was  the  rate  per  cent.  ? 

Give  case  11.     Analysis.     Hulc.     Case  III. 


PROBLEMS   IN   INTEREST.  245 

OPERATION.  Analysis.    Since  at 

$  4.50  1  per  cent.  $450,  in  3 

3  years,  -will  gain  $18. oO 

>^ ,  o  -  rt   .  interest,  the   rate   per 

S 1 3. OO,  int.  of  $150  for  3  years  at  1  percent.  ^      .      i  •  t    .i 

'  cent,  at  MJiicn  the  same 

$67.50  -^  13.50  =  5  i.ercent.,  Ans.         principal,  in  the   same 

time,  will  gain  $67.50, 
must  be  equal  to  the  number  of  times  $13.50  is  contained  in  $67.50; 
di\  iJing,  we  obtain  5,  the  required  rate  per  cent.     Hence  the 

Rule.  Divide  the  given  interest  hy  the  interest  on  the  prin- 
cipal for  the  given  time  at  1  per  cent.,  and  the  quotient  will  he 
the  rate  ptc^'  cent,  required. 

EXAMPLES    FOR    PRACTICE. 

2.  If  I  pay  S45  interest  for  the  use  of  $500  3  years, 
•what  is  the  rate  per  cent.  ?  Ans.    3. 

3.  The  interest  of  $180  for  1  year  2  months  6  days  is 
$12.78  ;  what  is  the  rate  per  cent.  ?  Ans.    6. 

4.  A  man  invests  $2000  in  bank  stock,  and  receives  a 
semi-annual  dividend  of  $75 ;    what  is  the  rate  jier   cent.  ? 

5.  At  what  per  cent,  must  $1000  be  loaned  for  3  years  3 
months  and  29  days,  to  gain  $183.18  ?  Ans.    5^, 

6.  A  man  builds  a  block  of  stores  at  a  cost  of  $21640,  and 
receives  for  them  an  annual  rent  of  $2596.80 ;  what  per  cent, 
does  he  receive  on  the  investment?  Ans.    12. 

CASE    IV. 

SIO.    Principal,  interest,  and  rate  per  cent.  Leing 

given,  to  find  the  time. 

1.  In  what  time  will  $360  gain  $86.40  intei'cst,  at  6  per 
cent.  ? 

orERATiON.  Analysis.     Since  in 

$  360  1  year  $360,  at  6  per 

.06  eent.,  will  gain  $21.60, 

the  number  of  years  in 
which  the  same  princi- 
pal, at  the  same  rate, 
will  gain  $86.40,  will  be 


$21.60     Iuterostof$3C0mlyearat6percent.      ,^|,ichthe    same    princi- 

$86.40  H- 21,60  =  4  years,  ylws.         pal,  at  the   same   rate, 


Analysis.     Rule.     Case  IV.     Analysis. 


2-i6  PERCENTAGE. 

as  many  as  $21.60  Is  contained  times  in  $86.40 ;  dividing,  we  ob- 
tLiin  4  years,  the  required  time.     Hence  the 

Rule.  Divide  the  given  interest  hy  the  interest  on  the  prin- 
cipal for  1  year,  and  the  quotient  will  he  the  time  required  in 
years  and  decimals. 

Note.  The  decimal  part  of  the  quotient,  if  any,  may  be  reduced  to 
months  and  days  (by  209). 

EXAMPLES    FOR    PRACTICE. 

2.  The  interest  of  $325  at  G  per  cent,  is  $58.50 ;  \vhat  is 
the  time  ?  j  Ans.    3  years. 

3.  B  loaned  $1G00  at  G  jier  cent,  until  it  amounted  to 
$2000  ;  what  was  the  time  ?  Ans.    4  years  2  months. 

4.  How  long  must  §204  be  on  interest  at  7  per  cent.,  to     g 
amount  to  $217.09  ?  Ans.    11  months. 

5.  Enc-a'Tinsr  in  business,  I  borrowed  $750  of  a  friend  at  6 

COO  ' 

per  cent.,  and  kept  it  until  it  amounted  to  $942  ;  how  long  did 
I  retain  it  ?  Ans.    4  years  3  months  G  days. 

G.  IIow  long  will  it  take  $200  to  double  itself  at  6  per  cent, 
simple  interest  ?  Ans.    1 G  years  8  months. 

7.    In  what  time  Avill  $G75  double  itself  at  5  per  cent.  ? 

Note.  The  time  in  years  in  which  any  sum  wiU  double  itself  may 
be  foimd  by  dividuig  100  by  the  rate  per  cent. 


co:\iPorxD  interest. 


31 


\l\.    Compound  Interest  is  interest  on  both  principal  and 
interest,  when  the  interest  is  not  paid  when  due. 

Note.  The  simple  interest  may  be  added  to  the  prmcipal  annually, 
semi-annually,  or  quarterly,  as  the  parties  may  agree  ;  but  the  taldng 
of  compound  interest  is  not  logal. 

1.    Wliat  is  the  compound  interest  of  $200,  for  3  years,  at 
G  ])tr  cent.? 

Rule.  In  what  time  wUl  any  sum  double  itself  at  interest  ?  "^Miat 
'fi  compound  interest  ? 


lii 


COMPOUND   INTEREST.  24.7 

OPERATION. 
$200  Principal  for  1st  year. 

$200  X  -06  =       12  Interest  for  1st  year. 

$212  Principal  for  2d  year. 

$212  X  -06  —       12.72      Interest  for  2d  year, 

$224.72      Principal  for  3d  year. 
$224.72  X  -06  =       13.483    Interest  for  3d  year. 

$238,203    Amount  for  3  years. 
200.000    Given  principal. 

$38,203    Compound  interest. 

Rule.  I.  Find  the  amount  of  the  given  princij^al  at  the 
given  rate  for  one  year,  and  make  it  the  -principal  for  the 
second  year. 

II.  Find  the  amount  of  this  new  principal,  and  make  it  the 
principal  for  the  third  year,  and  so  continue  to  do  for  the  given 
number  of  years.  ^ 

III.  Subtract  the  given  principal  from  the  last  amount,  and 
the  remainder  will  be  the  compound  interest. 

Notes.  1.  A^Tien  the  interest  is  payable  semi-annually  or  quar- 
terly, find  the  amount  of  the  given  principal  for  the  first  interval,  and 
make  it  the  principal  for  the  second  interval,  proceeding  in  all  respects 
as  •svhen  the  interest  is  payable  yearly. 

2.  When  the  time  contams  years,  months,  and  days,  find  the  amount 
for  the  years,  upon  which  compute  the  interest  for  the  months  and 
days,  and  add  it  to  the  last  amount,  before  subtracting. 

EXAMPLES    FOR    TKACTICE. 

2.  What  is  the  compound  interest  of  $500  for  2  years  at  7 
per  cent.  ?  Ans.    $72.45. 

3.  What  is  the  amount  of  $312  for  3  years,  at  6  per  cent, 
compound  interest  ?  Ans.    $371.59-|-. 

4.  What  is  the  compound  interest  of  $250  for  2  years, 
payable  semi-annually,  at  6  per  cent.?  Ans.    $31.37-f-. 

5.  What  will  $450  amount  to  in  1  year,  at  7  per  cent,  com- 
pound interest,  payable  quarterly  ?  Ans.    $482.33. 

6.  What  is  the  compound  interest  of  $236  for  4  years  7 
months  and  G  days,  at  G  per  cent.  ?  Ans.    $72.6G-j-. 

Explain  operation.     Give  rule. 


248 


PERCENTAGE. 


7.  "What  is  the  amount  of  $700  for  3  years  9  montlis  and 
24  days,  at  7  per  cent,  compound  interest  ?     Ans.    $906.55-|-. 

A  more  expeditious  method  of  computing  compound  interest 
than  the  preceding,  is  by  means  of  the  following 

TABLE, 

Showing  the  amount  0/"$!,  or  £1,  at  3,  4,  5,  6,  and  7  per  cent.,  compound 

interest,  for  any  number  of  years,  from  1  to  20. 


Yrs. 


1 

2 
3 
4 
5 


6 
7 
8 
9 
10 


11 
12 
13 
14 
15 


16 
17 
18 
19 
20 


3  per  cent. 


4  per  cent. 


5  per  cent. 


1.030,000  1.040,000  1.050,000 
1.060,900  1.081,600  1.102,500 
1.092,727  1.124,864  1.157,625 
1.125,509  1.109,859  1.215,506 


1.159,274  1.210,653 


1.194,052 
1.229,874 
1.206,770 
1.304,773 
1.343,916 


1.384,234 
1.425,761 
1.468,534 
1.512.590 
1.557,967 


1.265,319 
1.315,932 
1.368,569 


1.276,282 


1.340,096 
1.407,100 
1.477,455 


1.423,312  1.551,328 
1.480,244  1.628,895 


1.604,706 
1.652,848 
1.702,433 
1.753,506 


1.539,454  1.710,339 
1.601,032jl.795,856 
1.665,074  1.885,649 
1.731,67611.979,932 
1.800,044  ;2.078,928 


6  per  cent. 


1.060,000 
1.123,600 
1.191,016 
1.262,477 
1.338,226 


7  per  cent. 


1.418,519 
1.503,630 
1.593,848 
1.689,479 
1.790,848 


1.872,981  2.182,875 
1.947,900  2.292,018 
2.025,817|2.406,619 
2.106,849  2.526,950 


1.898,299' 
2.012,196' 
2.132,928 
2.260,904 
2.396,558 


1.806,111,2.191,123  2.653,298 


2.540,352 
2.692,773! 
2.854,339| 
3.025,600 
3.207,135 


1.07,000 
1.14,490 
1.22,504 
1.31,079 
1.40,255 

1.50,073 
1.60,578 
1.71,818 
1.83,845 
1.96,715 

2.10,485 
2.25,219 
2.40,984 
2.57,853 
2.75,903 

2.95,216 
!3.15,881 
3.37,293 
3.61,652 
3.86,968 


8.  AVhat  is  the  amount  of  $800  for  6  years,  at  7  per  cent. 

orERATION. 

From  the  table     $1.50073        Amount  of  $1  for  the  time. 

800     Principal. 

$1200.58400,     Ans. 

9.  What  is  the  compound  interest  of  $120  for  15  years,  at 
5  per  cent.?  Ans.    $129.47+. 


Of  what  use  is  the  table  in  computing  compound  interest  ? 


DISCOUNT.  249 

10.  "What  is  the  amount  of  $.10  fur  20  years,  at  7  per 
cent.?  Aks.    $.38 GOG. 

DISCOUNT. 

SI^.  Discount  is  an  abatement  or  allowance  made  for 
the  payment  of  a  debt  before  it  is  due. 

•Ilel.  The  Present  "Worth  of  a  debt,  pa^-able  at  a  future 
time  without  interest,  is  such  a  sum  as,  being  put  at  legal  in- 
terest, will  amount  to  the  given  debt  when  it  becomes  due. 

1.  A  owes  I>  $321,  payable  in  1  year;  what  is  the  pres- 
ent worth  of  the  debt,  the  use  of  money  being  Avorth  7  per 
cent.  ? 

OPERATiox.  Analysis.  The 

Am't  of     $1,  1.07  )  $321  ($300,  Present  value,  amount  of  $1    for 

321  1  year  is    $1.07; 

_  therefore  the  prcs- 

$321  Giveu  sum  or  debt.  ent  worth  of  every 

300  Present  worth.  $1.07  of  the  given 

$21  Discount.  fl'^^^t    is    $1;    and 

the  present  Avortli 
of  $321  will  be  as  many  dollars  as  $1.07  Is  contained  times  in  $321. 
$321  -^  1.07  =  $300,  Ans.     Hence  the  following 

KuLE.  I.  Divide  the  given  sum  or  debt  hy  the  amount  of 
$1  for  tlte  given  rate  and  time,  and  the  quotient  will  be  the  pres- 
ent worth  of  the  debt. 

II.  Subtract  the  present  worth  from  the  given  svm  or  debt, 
and  the  remainder  will  be  the  discount. 

Note.  The  terms  present  worth,  discount,  and  debt,  are  equivalent 
to  principal,  interest,  and  amount.  Hence,  Avhen  the  time,  rate  per 
cent.,  and  amount  are  given,  the  principal  may  be  found  by  (30§)  ; 
and  the  interest  by  subtracting  the  principal  from  the  amount. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  present  worth  of  $180,  payable  in  3  years 
4  months,  discounting  at  6  per  cent.  ?  Ans.    $150. 

Define  discount.     Present  worth.     Give  analysis.     Rule, 
11* 


250  PERCENTAGE. 

3.  What  is  the  present  worth  of  a  note  for  $1315.389,  due 
ill  2  years  6  months,  at  7  per  cent.?  Ans.    $1119.48. 

4.  What  is  the  present  worth  of  a  note  for  S8GG.038,  due 
in  3  years  6  months  and  G  days,  when  money  is  worth  8  per 
cent.  ?     What  the  discount  ?  Ans.    $190.15-}-,  discount. 

5.  What  is  the  present  worth  of  a  debt  for  §1005,  on  which 
$475  is  to  be  paid  in  10  months,  and  the  remainder  in  1  year  3 
months,  the  rate  of  interest  being  G  per  cent.  ? 

KoTE.  "When  payments  are  to  be  made  at  different  times  without 
interest,  find  the  present  worth  of  each  payment  separately,  and  take 
their  sum. 

Ans.   $94*.40-[-. 

G.  I  hold  a  note  against  C  for  $529,925,  due  Sept.  1,  1859 ; 
what  must  I  discount  for  the  payment  of  it  to-day,  Feb.  7, 
1859,  money  being  worth  G  per  cent.  ?  Ans.    $17,425. 

7.  A  man  was  offered  $3675  in  cash  for  his  house,  or 
$4235  in  3  years,  without  interest;  he  accepted  the  latter 
offej- ;  how  much  did  he  lose,  money  being  worth  7  per  cent.  ? 

Ans.    $175. 

8.  A  man,  having  a  span  of  horses  for  sale,  offered  them 
for  $480  cash  in  hand,  or  a  note  of  $550  due  in  1  year  8 
months,  without  interest ;  the  buyer  accepted  the  latter  offer  ; 
did  the  seller  gain  or  lose  thereby,  and  how  much,  interest  be- 
ing G  per  cent.  ?  Ans.  Seller  gained  $20. 

9.  What  must  be  discounted  for  the  present  payment  of  a 
debt  of  $2637.72,  of  which  $517.50  is  to  be  paid  in  6  months, 
$793.75  in  10  months,  and  the  remainder  in  1  year  G  months, 
the  use  of  money  being  worth  7  per  cent.  ?      Ans.  $187.29  -\-. 

10.  What  is  the  difference  between  the  interest  and  discount 
of  $130,  due  10  months  hence,  at  10  per  cent.  ?     Ans.  $.83^. 

PROMISCUOUS    EXAMPLES    IX    PERCENTAGE. 

1.  A  merchant  bought  sugar  in  New  York  at  6^  cents  per 
pound  ;  the  wastage  by  transportation  and  retailing  was  5  per 
cent.,  and  the  interest  on  the  first  cost  to  the  time  of  sale  was 
2  per  cent. ;  how  much  must  he  ask  per  pound  to  gain  25  per 
cent.?  Ami.   R'-f-  cents. 


PROMISCUOUS   EXAMPLES.  251 

2.  A  person  purchased  2  lots  of  land  for  $200  each,  and 
sold  one  at  40  per  cent,  more  than  cost,  and  the  other  at  20 
per  cent,  less  ;  how  much  did  he  gain  ?    •  A?is.    $40. 

3.  Sold  goods  to  the  amount  of  $425,  on  G  months'  credit, 
which  was  $25  more  than  the  goods  cost;  what  was  the  true 
profit,  money  being  worth  G  per  cent.  ?        Aiis.    $12.62  -f-. 

4.  Bought  cotton  cloth  at  13  cents  a  yard,  on  8  months' 
credit,  and  sold  it  the  same  day  at  12  cents  cash;  how  much 
did  I  gain  or  lose  per  cent.,  money  being  worth  6  per  cent.  ? 

Ans.    Lost  4  per  cent. 

5.  A  farmer  sold  a  pair  of  horses  for  $150  each;  on  one 
he  gained  25  per  cent.,  on  the  other  he  lost  25  per  cent. ;  did 
he  gain  or  lose  on  both,  and  how  much  ?       Ans.    Lost  $20. 

6.  A  man  invested  f  of  all  he  was  worth  in  the  coal  trade, 
and  at  the  end  of  2  years  8  months  sold  out  his  entire  interest 
for  $3100,  which  was  a  yearly  gain  of  9  per  cent,  on  the 
money  invested ;  how  much  was  he  worth  when  he  commenced 
trade?  Ans.    $3750. 

7.  In  how  many  years  Avill  a  man,  paying  interest  at  7  per 
cent,  on  a  debt  for  land,  pay  the  face  of  the  debt  in  interest  ? 

Ans.    14f  years. 

8.  Two  persons  engaged  in  trade ;  A  furnished  f  of  tlie 
capital,  and  B  f ;  and  at  the  end  of  3  years  4  months  they 
found  they  had  made  a  clear  profit  of  $5000,  which  was  12^- 
per  cent,  per  annum  on  the  money  invested  ;  how  much  cap- 
ital did  each  furnish  ?  Ans.   A,  $7500  ;  B,  $4500. 

9.  Bought  $500  wwth  of  dry  goods,  and  $800  worth  of 
groceries ;  on  the  dry  goods  I  lost  20  per  cent.,  but  on  the 
groceries  I  gained  15  per  cent. ;  did  I  gain  or  lose  on  the 
whole  investment,  and  hov/  much  ?  Ans.   Gained  $20. 

10.  What  amount  of  accounts  must  an  attorney  collect,  in 
order  to  pay  over  $1100,  and  retain  8^  per  cent,  for  collect- 
ing ?  Ans.    $1200. 

11.  A  merchant  sold  goods  to  the  amount  of  $667,  to  be 
paid  in  8  months ;  the  same  goods  cost  him  $600  one  year 
previous  to  the  sale  of  them ;  money  being  wortli  6  per  cent., 
what  was  his  true  gain  ?  Ans.    $5,346  -\-. 


252  PERCENTAGE. 

12.  A  nurseryman  sold  trees  at  $18  per  lumdred,  and 
cleared  ^  of  his  receipts ;   what  per  cent,  profit  did  he  make  ? 

Ans.    50  per  cent. 

13.  If  I  of  an  article  be  sold  for  what  f  of  it  cost,  what  is 
the  gain  per  cent.  ?  Atis.    40 g. 

14.  A  lumber  merchant  sells  a  lot  of  lumber,  which  he  has 
had  on  hand  G  months,  on  10  months'  credit,  at  an  advance  of 
30  per  cent,  on  the  first  cost ;  if  he  is  paying  5  per  cent,  inter- 
est on  capital,  what  are  his  profits  per  cent.  ?      A71S.    213.. 

15.  A  person,  owning  |  of  a  piece  of  property,  sold  20  per 
cent,  of  his  share  ;  what  part  did  he  then  own  ?        Ans.    i. 

IG.  A  speculator,  having  money  in  the  bank,  drew  GO  per 
cent,  of  it,  and  expended  30  per  cent,  of  50  per  cent,  of  this  for 
728  bushels  of  wheat,  at  $1.12i  per  bushel ;  how  much  was 
left  in  the  bank  ?  Ans.    $3G40. 

17.  I  wish  to  line  the  carpet  of  a  room,  tliat  is  G  yards  long 
and  5  yards  wide,  with  duck  |  yard  wide ;  how  many  yards  of 
lining  must  I  purchase,  if  it  will  shrink  4  per  cent,  in  length, 
and  5  per  cent,  in  width  ?  Ans.    43|f . 

18.  A's  money  is  28  per  cent,  more  than  B's  ;  how  many 
per  cent,  is  B's  less  than  A's  ?  Ans.    21^. 

19.  A  capitalist  invested  f  of  his  money  in  railroad  stock, 
which  depreciated  5  per  cent,  in  value ;  the  remaining  f  lie  in- 
vested in  bank  stock,  which,  at  the  end  of  1  year,  had  gained 
$1200,  which  was  12  per  cent,  of  the  investment ;  what  was  the 
whole  amount  of  his  capital,  and  what  was  his  entire  loss  or 
gain  ?  Ans.    $25000,  capital ;  $450,  gain. 

20.  C's  money  is  to  D's  as  2  to  3 ;  if  4-  of  C's  money  be 
put  at  interest  for  3  years  9  months,  at  10  per  cent.,  it  will 
amount  to  $1933.25 ;  how  much  money  has  each  ? 

Ans.   C,  $2812  ;  D,  $4218. 

BANKING . 
S14:.    A  Bailk  is  a  corporation  ciuirtered  by  law  for  the 
purpose  of  receiving  and  loaning  money,  and   furnisliing  a 
paper  circulation. 


"Wlmt  is  a  bank  ? 


BANKING.  253 

315.  A  Promissory  Note  is  a  written  or  printed  engage- 
ment to  pay  a  certain  sum,  either  on  demand  or  at  a  specified 
time. 

38 60  Bank  Notes,  or  Bank  Bills,  are  the  notes  made 
and  issued  by  banlis  to  circuhxte  as  money.  They  are  joayable 
in  specie  at  the  banks. 

SB  17.  Tlie  Face  of  a  note  is  the  sum  made  payable  by 
the  note. 

?II8.  Days  of  Grace  are  the  three  days  usually  allowed 
by  law  for  the  payment  of  a  note  after  the  expiration  of  the 
time  specified  in  the  note. 

Si9.  The  Matui'ity  of  a  note  is  the  expiration  of  the 
days  of  grace  ;  a  note  is  due  at  maturity. 

S5®.  Notes  may  contain  a  promise  of  interest,  which  will 
be  reckoned  from  the  date  of  the  note,  unless  some  other  time 
be  specified. 

The  transaction  of  borrowing  money  at  banks  is  conducted 
in  accordance  with  the  following  custom  :  the  borrower  pre- 
sents a  note,  either  made  or  indorsed  by  himself,  payable  at 
a  specified  time,  and  receives  for  it  a  sum  equal  to  the  face, 
less  the  interest  for  the  time  the  note  has  to  run.  The  amount 
thus  withheld  by  the  bank  is  in  consideration  of  advancing 
money  on  the  note  prior  to  its  maturity. 

3^21.  Bank  Discount  is  an  allowance  made  to  a  bank  for 
the  payment  of  a  note  before  it  becomes  due. 

JI2ilo  The  Proceeds  of  a  note  is  the  sum  received  for  it 
when  discounted,  and  is  equal  to  the  face  of  the  note  less  the 
discount. 

CASE    I. 

S?I3.    Given  the  face  of  a  note  to  find  the  proceeds. 
The  law  of  custom  at  banks  makes  the  discount  of  a  note 

Define  a  promissory  note.  Banit  notes.  Tlie  face  of  a  note.  Days 
of  grace.  The  maturity  of  a  note.  Explain  the  process  of  discounting 
a  note  at  a  bank.  Defijie  bank  discoixnt.  The  proceeds  of  a  note. 
What  is  Case  I  ? 


254:  PERCENTAGE. 

equcil  to  tlie  simple  interest  at  the  legal  rate  for  the  time  spe- 
cified in  the  note.     Hence  the 

Role.  I.  Compute  the  interest  on  the  face  of  the  note  for 
three  days  more  than  the  specified  time  ;  the  result  will  he  the 
discount. 

II.  Subtract  the  discount  from  the  face  of  the  note,  and  the 
remainder  will  he  the  proceeds. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  discount,  and  what  the  proceeds,  of  a  note 
for  $450,  at  60  days,  discounted  at  a  bank  at  6  per  cent.  ? 

Ans.    Discount,  $4,725  ;  proceeds,  $445,275. 

2.  What  are  the  proceeds  of  a  note  for  $368,  at  90  days, 
discounted  at  the  Bank  of  New  York  ?      Ans.    $361,345  -f-. 

3.  What  shall  I  receive  on  my  note  for  $475.50,  at  60 
days,  if  discounted  at  the  Crescent  City  Bank,  New  Orleans  ? 

Aus.    $471.33-1-. 

4.  What  are  the  proceeds  of  a  note  for  $10000,  at  90  days, 
discounted  at  the  Philadelphia  Bank  ?  Ans.    $984.5. 

5.  Paid,  in  cash,  $240  for  a  lot  of  merchandise.  Sold  it 
the  same  day,  receiving  a  note  for  $250  at  60  days,  which  I 
20t  discounted  at  the  Hartford  Bank.  What  did  I  make  by 
this  speculation  ?  Ans.    ^t.oi^. 

6.  A  note  for  $360.76,  drawn  at  90  days,  is  discounted  at 
the  Vermont  Bank.     Find  the  proceeds.      Ans.   $355.168  4-. 

7.  Wishing  to  borrow  $530  of  a  Avestern  bank  which  is 
discounting  paper  at  8  per  cent.,  I  give  my  note  for  $536.75, 
payable  in  60  days.  How  much  do  I  need  to  make  up  the 
required  amount  ?  Ans.    $.7645. 

Notes.  1.  To  indicate  the  matiiritj'  of  a  note  or  draft,  a  vortical 
line  (  I  )  is  used,  with  the  day  at  which  the  note  is  iiominally  due  ou 
the  left,  and  the  date  of  maturity  on  the  right ;  thus,  Jan.  '  |  m  • 

2.  ^\^■lcn  a  note  is  on  interest,  payable  at  a  future  specified  time,  the 
amount  is  the  face  of  the  note,  or  th.e  sum  made  payable,  and  must  be 
made  the  basis  of  discount. 


Give  rule. 


BANKING.  255 

Find  the  maturity,  term  of  discount,  and  proceeds  of  the 
following  notes  :  — 


=i^'>'"^•  Boston,  Jan.  4,  1859. 

^s.  Three  months  after  date,  I  promise  to  pay  to  the  order  of 
John  Brown  &  Co.  five  hundred  dollars,  at  the  Suffolk  Bank, 
value  received.  James  Barker. 

Discounted  March  2.  (Due,  April*!   . 

Ans.    \  Terra  of  discount,  3G  da. 
Proceeds,  §497. 


^"•^^'-  St.  Louis,  June  12,  1859. 

9.  Six  months  after  date,  I  promise  to  pay  Thomas  Lee,  or 
order,  seven  hundred  fifty  dollars,  with  interest,  value  re- 
ceived. ,  Byron  Quinby. 

Discounted  at  a  broker's,  Nov.  15,  at  10  per  cent. 


Due,  Dec. 


121 


1  5' 


Ans.    {  Term  of  discount,  30  da. 
Proceeds,    $7GG.434+. 

CASE    II. 

3S4.  Given  the  proceeds  of  a  note,  to  find  the 
face. 

1.  I  wish  to  borrow  $1:00  at  a  bank.  For  what  sum  must 
I  draw  my  note,  payable  in  60  days,  so  that  when  discounted 
at  6  per  cent.  I  shall  receive  the  desired  amount  ? 

OPERATION.  Analysis.     $400  is  the 

SLOOOO  proceeds  of  a  certain  note, 

.0105  r=  disc,  on  $1  for  63  da.  the  face   of  which  Ave   are 

required  to  find.    We  first 

$  .9895  =:  proceeds  of  $1.  obtain  the  proceeds  of  ,'^1 
$400  -i-  .9895  r=  $404,244  =  by  the  last  case,  and  then 
face  of  the  required  note.  divide  the  given  proceeds, 
$400,  by  this  sum  ;  for,  as  many  times  as  the  proceeds  of  $1  is  con- 
tained in  the  given  proceeds,  so  many  dollars  must  be  the  face  of 
the  required  note.     Hence  the 


Give  Case  II.     Analysis. 


256  PERCENTAGE. 

KuLE.  Diride  the  jiroceeds  hy  the  proceeds  of  $1  for  the 
time  and  rate  mentioned,  and  the  quotient  toill  he  the  face  of 
ilic  note. 

EXAMPLES    FOll    FRACTICE. 

2.  What  is  the  face  of  a  note  at  GO  days,  which  yields 
$680  when  discounted  at  a  New  Haven  bank  ? 

Ans.    $687,215+. 

3.  What  is  the  face  of  a  note  at  90  days,  of  which  the  pro- 
ceeds are  $1000  when  discounted  at  a  Louisiana  bank  ? 

Ans.    $1013.085  -[-. 

4.  Wishing  to  borrow  $500  at  a  bank,  for  what  sura  must 
my  note  be  drawn,  at  30  days,  to  obtain  the  required  amount, 
discount  being  at  7  per  cent.  ?  Ans.    $503.22  -{-• 

5.  James  Hopkins  buys  merchandise  of  me  in  New  York, 
at  cash  price,  to  the  amount  of  $1256.  Not  having  money, 
he  gives  his  note  in  payment,  drawn  at  6  months.  What 
must  be  the  foce  of  the  note  ?  Ans.   $1302.341  +. 


EXCIL.\NGE. 

S^«>.  Exchange  is  a  method  of  remitting  money  from  one 
place  to  another,  or  of  making  payments  by  written  orders. 

3'26.  A  Bill  of  Exchange  is  a  written  request  or  order 
upon  one  person  to  pay  a  certain  sum  to  another  person,  or  to 
his  order,  at  a  specified  time. 

327.  A  Sight  Draft  or  Bill  is  one  requiring  payment  to 
be  made  "at  sight,"  which  means,  at  the  time  of  its  presenta- 
tion to  the  person  ordered  to  pay.  In  other  bills,  the  time 
specified  is  usually  a  certain  number  of  days  "after  sight." 

There  are  always  three  parties,  and  usually  four,  to  a  trans- 
action in  exchange : 

II2S.  The  Drawer  or  Maker  is  the  person  who  signs  the 
order  or  bill. 

Give  the  rule.  Define  exchange.  A  bill  of  exchange.  A  sight  draft. 
The  diawer. 


EXCHANGE.  257 

3^9.  The  Drawee  is  the  person  to  whom  the  order  is 
addressed. 

SJI®.  The  Payee  is  the  person  to  whom  the  money  is  or- 
dered to  be  paid. 

5511 .  The  Buyer  or  Remitter  is  the  person  who  purchases 
the  bin.  lie  may  be  himself  tlie  payee,  or  the  bill  may  be 
drawn  in  favor  of  any  other  person. 

5512.  The  Indorsement  of  a  bill  is  the  writing  upon  its 
back,  by  which  the  payee  relinquishes  his  title,  and  transfers 
the  payment  to  another.  The  payee  may  indorse  in  blank  by 
writing  his  name  only,  which  makes  the  bill  payable  to  the 
hearer,  and  consequently  transferable  like  a  bank  note  ;  or  he 
may  accompany  his  signature  by  a  special  order  to  pay  to 
another  person,  who  in  his  turn  may  transfer  the  title  in  like 
manner.  Indorsers  become  separately  responsible  for  the 
amount  of  the  bill,  in  case  the  drawee  fails  to  make  payment. 
A  bill  made  payable  to  the  hearer  is  transferable  without  in- 
dorsement. 

330.  The  Acceptance  of  a  bill  is  the  promise  which  the 
drawee  makes  when  the  bill  is  presented  to  him  to  pay  it  at 
maturity  ;  this  obligation  is  usually  acknowledged  by  writing 
the  word  "  Accepted,"  with  his  signature,  across  the  face  of 
the  bill. 

Note.  Three  days  of  ^ace  are  usually  allowed  for  the  pa;(Tnent  of 
a  l)ill  of  exchange  after  the  time  specified  has  expii-ed.  But  in  New 
York  State  no  grace  is  allowed  on  sight  drafts. 

From  these  definitions,  the  use  of  a  bill  of  exchansfe  in  mon- 
etary  transactions  is  readily  perceived.  If  a  man  wishes  to 
make  a  remittance  to  a  creditor,  agent,  or  any  other  person 
residing  at  a  distance,  instead  of  transporting  specie,  which  is 
attended  wi-rii  expense  and  risk,  or  sending  bank  notes,  which 
are  liable  to  be  uncurrent  at  a  distance  from  the  banks  that 
issued  them,  he  remits  a  bill  of  exchange,  purchased  at  a  bank 
or  elsewhere,  and  made  payable  to  the  proper  person  in  or 

The  drawee.  The  payee.  The  buyer.  An  indorsement.  An 
acceptance.     What  of  grace  on  bills  of  exchange  ? 


25S  PERCENTAGE. 

near  the  place  where  he  resides.  Thus  a  man  by  paying 
Boston  funds  in  Bctton,  may  put  New  York  funds  into  tlie 
hands  of  his  New  York  agent. 

SS4.  Tlie  Course  of  Exchange  is  the  variation  of  the 
cost  of  sight  hills  from  their  par  value,  as  affected  by  the  rela- 
tive conditions  of  trade  and  commercial  credit  at  the  two  places 
between  which  exchange  is  made.  It  may  be  either  at  a  pre- 
mium or  discount,  and  is  rated  at  a  certain  per  cent,  on  the 
face  of  tlie  bill.  Bills  payable  a  specified  time  after  siglit  are 
subject  to  discount,  like  notes  of  hand,  for  the  term  of  credit 
given.  Hence  their  value  in  the  money  market  is  affected  by 
both  the  course  of  exchange  and  the  discount  for  time. 

8S*>.  Foreign  Exchange  relates  to  remittances  made  be- 
tween different  countries. 

S3€.  Domestic  or  Inland  Exchange  relates  to  remit- 
tances made  between  diflerent  places  in  the  same  country. 

An  inland  bill  of  exchange  is  commonly  called  a  Draft. 
In  this  work  we  shall  treat  only  of  Inland  Exchange. 

CASE   I. 

SSr.    To  find  the  cost  of  a  draft. 


$500.  Syracuse,  May  7,  1859. 

1.    At  sight,  pay  to  James  Clark,  or  order,  five  hundred 
dollars,  value  received,  and  charge  the  same  to  our  account. 
To  M.  Smith  &  Co. 

Messrs.  Brown  «&;  Foster,  \ 
Baltimore.         ^ 
What  is  the  cost  of  tlie  above  draft,  the  rate  of  exchange 
being  1^  per  cent,  premium? 

operation.  Analysis.     Since  cx- 

$500  X  1 .0 1 5  r=  $507.50,  Ans.         ^^'"^"^  ^«  "^  H  Fj;  c^'»t 

premium,  each  dollar  ot 

the  draft  Avill  cost  $1,015  ;  and  to  find  the  whole  cost  of  the  draft, 

How  is  exchange  conducted?  Explain  coiirse  of  cxchanc:e.  For- 
cis;n  exchange.  Inland  exchange.  Define  a  draft.  "What  is  Case  I  ? 
Give  analysis. 


EXCHANGE.  259 

we  multiply  its  face,  $.300,  by  1,015,  and  obtain  $507.50,  the  re- 
quired Ans. 

$480.  Boston,  June  12,  1859. 

2.  Thirty  days  after  sight,  pay  to  John  Otis,  or  bearer,  four 
hundred  eiglity  dollars,  value  received,  and  charge  the  same 
to  account  of  Amos  Tkenchard. 

To  John  Stiles  &  Co.,  ) 
New  York. ) 

What  is  the  cost  of  the  above  draft,  exchange  being  at  a 
premium  of  3  per  cent.  ? 

OPERATION.  Analysis.      Since 

$1.0000  time    is     allowed,    the 

.OOoa  :r=  discount  for  33  days.  di'aft   must   suflar   dis- 

count in  the  sale.     The 


$    .9045  =r  proceeds  of  SI. 
.03        ^z  rate  of  exchange. 


discount  of  $1,  at  the 
legal  rate  in  Boston,  for 
$1.0245  =  cost  of  $1  of  the  draft.  the   specified   time,   al- 

$480  X  1.0245  =  $491.76,  Ans.  lowing  grace,  is  $.0055, 

which,  sulitracted  from 
$1,  gives  $.9945,  the  cost  of  $1  of  the  draft,  provided  sight  ex- 
change were  at  par  ;  but  sight  exchange  being  at  preiuiiim,  we  add 
the  rate,  .03,  to  .9945,  and  obtain  $1.0245,  the  actual  cost  of  $1. 
Then,  multiplying  $480  by  1.0245,  we  obtain  $491.70,  the  Ans. 
From  these  examples  we  derive  the  following 

Rule.  I.  For  sight  drafts.  —  Ilultipli/ the  face  of  the  draft 
hfl  1  2}lus  the  rate  xvhen  exchange  is  at  a  premium,  and  hj 
1  minus  the  rate  lohen  exchange  is  at  a  discount. 

II.  For  drafts  payable  after  sight.  —  Find  the  proceeds  of  $1 
at  hank  discount  for  the  specified  time,  at  the  legal  rate  where 
the  draft  is  purchased ;  then  add  the  rate  of  exchange  tvhen 
at  a  premium,  or  subtract  it  when  at  a  discount,  and  multiphj 
the  face  of  the  draft  hy  this  result. 

EXAMPLES    FOR   PRACTICE. 

3.    A  mercha,nt  in   Cincinnati  wishes  to  remit  $1000   by 
Give  analysis,    Rule  I;  II. 


260  PERCENTAGE. 

draft  to  his  agent  in  New  York  ;  what  will  the  bill  cost,  ex- 
change being  at  3  per  cent,  premium?  Ans.    $1030. 

4.  What  will  be  the  cost  in  Rochester  of  a  draft  on  Albany 
for  $400,  payable  at  sight,  exchange  being  at  f  per  cent,  pre- 
mium ?  'Ans.    $403. 

5.  A  merchant  in  St.  Louis  orders  goods  from  Kew  York, 
to  the  amount  of  $530,  which  amount  he  remits  by  draft,  ex- 
tliange  being  at  2|  per  cent,  premium.  If  he  pays  $20  for 
transportation,  what  will  the  goods  cost  him  in  St.  Louis  ? 

Ans.    $564,575. 
G.    What  will  be  the  cost,  in  Detroit,  of  a  draft  on  Boston 
for  $800,  payable  GO  days  after  sight,  exchange  being  at  a  pre- 
mium of  2  per  cent.?  Ans.    $806.20. 

7.  A  man  in  Philadelphia  purchased  a  draft  on  Chicago  for 
$420,  payable  30  days  after  sight ;  what  did  it  cost  him,  the 
rate  of  exchange  being  1  J- per  cent,  discount?    Ans.  $411.39. 

8.  A  merchant  in  Portland  receives  from  his  agent  320 
barrels  of  floui',  purchased  in  Chicago  at  $10  per  barrel ;  in 
I)ayraent  for  which  he  remits  a  draft  on  Chicago,  at  2|-  per 
cent,  discount.  Tlie  transportation  of  his  flour  cost  $312. 
What  must  he  sell  it  for  per  barrel  to  gain  $400  ?     Ans.  $12. 

CASE    11. 

338.  To  find  tlio  face  of  a  draft  which  a  given  sum 
will  purchase. 

1.    A  man  in  Indiana  paid  $3G9.72  for  a' draft  on  Boston, 

drawn  at  30  days  ;  what  Avas  the  face  of  the  draft,  exchange 

being  at  3^  per  cent,  premium? 

OPERATION.  Analysis.     We  find, 

$3G9.72  —  1.027  =  $360,  Ajis.        ^'y  <^^^^  ^'  t^"*  ^  '^^""^^ 
■  '  for  $1  will  cost  $1,027; 

hence  the  draft  that  will  cost  $309.72  must  be  for  as  many  dollars  as 

1.027  is  contained  times  in  $360.72  ;  dividinp:,  we  obtain  $300,  the 

Ans.     From  this  example  and  analysis  we  derive  the  following 

"NMiat  is  Case  II  ?     Give  analysis. 


EQUATION   OF   PAYMENTS.  261 

Rule.  Divide  the  given  cost  by  the  cost  of  a  draft  for  $1, 
at  the  given  rate  of  exchange  ;  the  quotient  ivill  he  the  face  of 
the  required  draft. 

EXAMPLES    FOR    PRACTICE. 

2.  What  draft  may  be  purchased  for  $243. GO,  exchange 
being  at  1^  per  cent,  premium?  Ans.    $2-iO. 

3.  What  draft  may  be  purchased  for  $79.20,  exchange  be- 
ing at  1  per  cent,  discount  ?  ^  Ans.    $80. 

4.  An  agent  in  Pittsburg  holding  $282.66,  due  his  em- 
ployer in  New  Haven,  is  directed  to  malvc  tlie  remittance 
by  draft,  drawn  at  60  days.  What  will  be  the  face  of  the 
draft,  exchange  being  at  2  per  cent,  premium  ?       Ans.    S280. 

5.  An  emigrant  from  Bangor  takes  $240  in  bank  bills  to 
St.  Paul,  Min.,  and  there  pays  4-  per  cent,  brokerage  in  ex- 
change for  current  money.  What  would  he  have  saved  by 
purchasing  in  Bangor  a  draft  on  St.  Paul,  drawn  at  30  days, 
exchange  being  at  1^  per  cent,  discount  ?        Ayis.   $5.599 -[-. 

6.  A  Philadelphia  manufacturer  is  informed  by  his  agent  in 
Buffalo  that  $3600  is  due  him  on  the  sale  of  some  property. 
He  instructs  the  agent  to  remit  by  a  draft  payable  in  GO  days 
after  sight,  exchange  being  at  f  per  cent,  premium.  The  agent, 
by  mistake,  remits  a  sight  draft,  which,  when  received  in  Phila- 
delphia, is  accepted,  and  paid  after  the  expiration  of  the  three 
days  of  grace.  If  the  manufacturer  immediately  puts  this 
money  at  interest  at  the  legal  I'ate,  will  he  gain  or  lose  by  the 
blunder  of  his  agent  ?  Ans.   He  will  lose  $8.24-j-. 

EQUATION   OF   PAYMENTS. 

«l«IO.  Equation  of  Payments  is  the  process  of  finding  the 
mean  or  equitable  time  of  payment  of  several  sums,  due  at 
different  times  without  interest. 

S-l®.  The  Term  of  Credit  is  the  time  to  elapse  before  a 
debt  becomes  due. 

Rule.     Define  epilation,  of  payments.     Term  of  credit. 


262  EQUATION    OF    PAYMENTS. 

?I41.  The  Average  Term  of  Credit  is  the  time  to  elapse 
before  several  debts,  due  at  different  times,  may  all  be  paid 
at  once,  witliout  loss  to  debtor  or  creditor. 

S49.  The  Equated  Time  is  the  date  at  which  the  several 
debts  may  be  canceled  by  one  payment. 

CASE    I. 

343,  Wlien  all  the  terms  of  crcclLt  begin  at  the 
game  date. 

1.  On  the  first  day  of  January  I  find  that  I  owe  Mr.  Smith 
8  dollars,  to  be  paid  in  5  months,  10  dollars  to  be  paid  in  2 
months,  and  12  dollars  to  be  paid  in  10  months;  at  what  time 
may  I  pay  the  whole  amount  ? 

OPERATIOX. 
$    8  X      5  zr:     40 

10  X    2  ==    20 
22  X  10  =  120 

30  180  -^  30  =:  6  mo.,  average  time  of  credit. 

Jan.  1.  -f-  6  mo.  =  July  1,  equated  time  of  payment. 
Analysis.  The  whole  amount  to  be  paid,  as  seen  above,  is  830 : 
and  we  are  to  find  how  long  it  shall  be  withheld,  or  what  term  of 
credit  it  shall  have,  as  an  equivalent  for  the  various  terms  of  credit 
on  the  different  items.  Now,  the  value  of  credit  on  any  sum  is  meas- 
ured by  the  product  of  the  money  and  time.  And  we  say,  the  credit 
on  $8  for  5  mo.r=the  credit  on  $40  for  1  mo.,  because  8  X  ^  =  40 
X  1.  la  the  same  manner,  Ave  have,  the  credit  on  $10  for  2  mo.= 
the  credit  on  $20  for  1  mo. ;  and  the  credit  on  $12  for  lOmo.m: 
the  credit  on  $120  for  1  mo.  Hence,  by  addition,  the  value  of  the 
several  terms  of  credit  on  their  respective  sums  equals  a  credit  of  1 
month  on  $180;  and  this  equals  a  credit  of  6  months  on  $30,  be- 
CLiuse 

30  X  6  r=  180  X  1. 

Rule.  I.  Multiply  each  payment  by  its  term  of  credit,  and 
divide  fho  sum  of  the  products  hj  the  sum  of  the  payments  ;  the 
quotient  will  he  tha  average  term  of  credit, 

Average  term  of  credit.     Equated  time.     GiVe  Case  I.     Analvsis. 
Kulc. 


AVERAGING   CREDITS.  263 

II.  Add  the  average  term  of  credit  to  the  date  at  which  all 
the  credits  begin,  and  the  result  will  he  the  equated  time  of 
payment. 

Notes.  1.  The  periods  of  time  used  as  multipliers  must  all  be  of 
the  same  denomination,  and  the  quotient  will  be  of  the  game  denomi- 
nation as  the  terms  of  credit ;  if  these  bo  months,  and  there  1)0  a  re- 
mainder after  the  division,  contiiuie  the  division  to  days  by  reduction, 
always  taking  the  nearest  unit  in  the  last  result. 

2.  The  several  rules  in  equation  of  payments  are  based  upon  the 
principle  of  bank  discount ;  for  they  imply  that  the  discount  of  a  sum 
paid  before  it  is  due  equals  the  interest  of  the  same  amount  paid  after 
it  is  due. 

EXAMPLES    FOR    PRACTICE. 

2.  On  the  25th  of  September  a  trader  bought  merchandise, 

as  follows  :  $700  on  20  days'  credit ;  $400  on  30  days'  credit ; 

$700  on  40  days'  credit:  what  was  the  average  term  of  credit, 

and  what  the  equated  time  of  payment  ? 

,        (  Average  credit,  30  days. 
Ans.  ■{  ^  '  ■^       ^      «K 

(.  Equated  time  of  payment,  Oct.  2o. 

3.  On  July  1  a  merchant  gave  notes,  as  follows :  the  first 
for  $250,  due  in  4  months ;  the  second  for  $750,  due  in  2 
months;  the  third  for  $500,  due  in  7  months:  at  what  time 
may  they  all  be  paid  in  one  sum  ?  Ans.    Nov.  1. 

4.  A  farmer  bought  a  cow,  and  agreed  to  pay  $1  on  Mon- 
day, $2  on  Tuesday,  $3  on  Wednesday,  and  so  on  for  a  week  ; 
desirous  afterward  to  avoid  the  Sunday  payment,  he  offered  to 
pay  the  whole  at  one  time :  on  what  day  of  the  week  would 
this  payment  come  ?  Ans.    Friday. 

5.  Jan.  1,  I  find  myself  indebted  to  John  Kennedy  in  sums 
as  follows  :  $G50  due  in  4  months ;  $725  due  in  8  months  ;  and 
$500  due  in  12  months  :  at  what  date  may  I  settle  by  giving 
my  note  on  interest  for  the  whole  amount?    Ans.  Aug.  21. 

CASE  ir. 

S44.  When  tlio  terms  of  credit  begin  at  different 
dates,  and  tlie  account  has  only  one  side. 

H'lii'S.  An  Account  is  the  statement  or  record  of  mercantile 
transactions  in  business  form. 

Give  Case  II.     Define  an  accotint. 


264: 


EQUATION   OF   PAYMENTS. 


S-16.  The  Items  of  an  account  may  be  sums  due  at  the 
date  of  the  transaction,  or  on  credit  for  a  specified  time. 

An  account  may  have  both  a  debit  and  a  credit  side,  the 
former  marked  Dr.,  the  latter  Cr.  Suppose  A  and  B  have 
dealings  in  which  there  is  an  interchange  of  money  or  prop- 
erty ;  A  keeps  the  account,  heading  it  with  B's  name ;  the  Dr. 
side  of  the  account  shows  what  B  has  received  from  A ;  the 
Cr.  side  shows  what  he  has  parted  with  to  A. 

SJiT.  The  Balance  of  account  is  the  difference  of  the  two 
sides,  and  may  be  in  favor  of  either  party. 

If,  in  the  transactions,  one  party  has  received  nothing  from 

the  other,  tlie  balance  is  simply  the  whole  amount,  and  the 

account  has  but  one  side.     Bills  of"  purchase  are  of  this  class. 

Note.     Book  accounts  bear  interest  after  the  expiration  of  the  term 
of  credit,  and  notes  after  thoy  become  due, 

348.  To  Average  an  Account  is  to  find  the  mean  or 
equitable  time  of  payment  of  the  balance. 

349.  A  Focal  Date  is  a  date  to  which  all  the  others  are 
compared  in  averaging  an  account. 

1.  When  does  the  amount  of  the  following  bill  become  due, 
by  averaging  ? 

J.  C.  Smith, 

1859.                      To  C.  E.  BoEDEN,       Dr. 
June    1.     To  Cash, $450 

"     12.      "    Mdse.  on  4  mos., 500 

Aug.  IG.      "    Mdse., 250 


FIRST   OPERATION. 


SECOND   OPERATION. 


Due. 

da. 

Items. 

Prod. 

Jime  1 

Oct.  12 

i  Aug.  16 

0 

133 

76 

450 
500 
250 

1200 

66500 
19000 

85500 

85500-1-1200  =  71  da. 
.       ^11  da.  after  June  1, 
^^*-  \  or  Aug.  11. 


Due. 

da. 

Items. 

Prod. 

June  1 
Oct.  12 

Aug.  16 

133 

.0 

57 

450 
500 
250 

59850 

14250 
74100 

1200 

74100 -I- 1200  r=  62  da. 
.       5  62  da.  before  Oct.  12, 
^"*-^  or  Aug.  11. 


Define  items.     Balance. 


To  average  an  account. 


A  focal  date. 


i 


AVERAGING    ACCOUNTS.  265 

Analysis.  By  reference  to  the  example,  it  will  be  seen  that  the 
items  are  due  June  1,  Oct.  12,  and  Aug.  16,  as  shown  in  the  two 
operations.  In  the  first  operation  we  use  the  earliest  date,  June  1, 
as  a  focal  date,  and  find  the  difference  in  days  between  this  date  and 
each  of  the  others,  regard  being  had  to  the  number  of  days  in  cal- 
endar months.  From  June  1  to  Oct.  12  is  133  da. ;  from  June  1  to 
Aug.  16  is  7G  da.  Hence  the  first  item  has  no  credit  from  June  1, 
the  second  item  has  133  days'  credit  from  June  1,  and  the  third 
item  has  76  days'  credit  from  June  l,*as  appears  in  the  column 
marked  da.  After  this  we  proceed  precisely  as  in  Case  I,  and  find 
the  average  credit, 71  da.,  and  the  equated  time,  Aug.  11. 

In  the  second  operation,  the  latest  date,  Oct.  12,  is  taken  for  a 
focal  date  ;  the  work  is  explained  thus :  Suppose  the  account  to  be 
settled  Oct.  12.  At  that  time  the  first  item  has  been  due  133  days, 
and  must  therefore  draw  interest  for  this  time.  But  interest  on 
$450  for  133  days  =  the  interest  on  $598o0  for  1  da.  The  second 
item  draws  no  interest,  because  it  falls  due  Oct.  12.  The  third  item 
must  draw  interest  57  days.  But  interest  on  $250  for  57  days  = 
the  interest  on  $14250  for  1  day.  Taking  the  sum  of  the  products, 
we  find  the  whole  amount  of  interest  due  on  the  account,  at  Oct.  12, 
equals  the  interest  on  $74100  for  1  day ;  and  this,  by  division,  is 
found  to  be  equal  to  the  interest  on  $1200  for  62  days,  which  time 
is  the  average  term  of  interest.  Hence  the  account  would  be  settled 
Oct.  12,  by  paying  $1200  with  interest  on  the  same  for  62  days.  This 
shows  that  1200  has  been  due  62  days  ;  that  is,  it  falls  due  Aug.  11, 
without  interest.     Hence  the  following 

EuLE.  I.  Find  the  time  at  lohich  each  item  becomes  due, 
by  adding  to  the  date  of  each  transaction  the  term  of  credit,  if 
any  be  specified,  and  write  these  dates  in  a  column, 

II.  Assume  either  the  earliest  or  the  latest  date  for  a  faced 
date,  and  find  the  difference  in  days  between  the  focal  date  and 
each  of  the  other  dates,  and  write  the  results  in  a  second  column. 

III.  Write  the  items  of  the  account  in  a  third  column,  and 
multiply  each  sum  by  the  corresponding  number  of  days  in  the 
preceding  column,  writing  the  products  in  a  final  column. 

IV.  Divide  the  sum  of  the  products  by  the  sum  of  the  items. 
The  quotient  will  be  the  average  term  of  credit  when  the 
— — ■ • — ^ — . _-. — . —  ) 

Give  analysis.    Rule. 
RP.  12 


266  EQUATION   OF  PAYMENTS.  , 

earliest  date  is  the  focal  date,  or  the  average  term  of  interest 
when  the  latest  date  is  the  focal  date ;  in  either  case  always 
reclcon  from  the  focal  date  toicard  the  other  dates,  to  find  the 
equated  tune  of  payment. 

examples  for  practice. 

2.  John  Brown, 

1859c,  To  James  Greigg,      Dr. 

Jan.     1.     To      50  yds.  Broadcloth,  (d)  $3.00,  ...  $150 
"      IG.      "   2000    "     Calico,  "       .10,...     200 

Feb.     4.      "        75    "     Carpeting,     «      1.33^,,.     100 
March  3.      «      400    "     Oil  Cloth,     "       .40,  ...     IGO 
If  James  Greigg  wishes  to  settle  the  above  bill  by  giving 
his  note,  from  what  date  shall  the  note  di-aw  interest  ? 

Ans.   Jan.  27. 

3.  Adram  Russel, 

1859  To  Wtnkoop  &  Brc,    Dr. 

March  lo  To  Cash,  ..,  c  r  c  o  r  » ..,.«..  -^  ...  $300 

April    4.  "    Mdse.,  .cooo.o,„ooococcc.-.    240 

June  18.  "        "       on  2  mo.,  ooooooc-oc  c  ^    100 

What  is  the  equated  time  of  payment  of  the  above  account  ? 

Ans.   May  20. 

4.  John  Otis, 

1858.  To  James  Ladd,       Dr. 

June  1.       To  500  bu.  Wheat,  Co)  $1.20, .  .  ,  - . .  $G00 

1.50 „    300 

1.30,  u .  .  ■  .  n  8o2 
1.00,..  .„,,..    7G0 

l.OU,  p  •  o  c  •  •      i  0\} 

When  is  the  whole  amount  of  the  above  bill  due,  per 
average?  Ans.   June  18. 

5.  My  oxponditnres  in  building  a  house,  in  tlie  year  1856, 
were  as  follows  :  Jan.  1  G,  $530.78  ;  Feb.  20,  $  125.30  ;  JMarch  4, 
$259.25 ;  April  24,  $78G.3G.     If  at  the  last  date  I  agree  to 


(( 

12. 

(( 

200 

a 

n 

li 

Ci 

15. 

(( 

040 

li 

u 

ii 

(( 

25. 

tk 

700 

ii 

u 

ii 

(( 

30. 

t( 

500 

u 

ii 

ii 

AVERAGING   ACCOUNTS. 


267 


sell  the  house  for  exactly  what  it  cost,  with  reference  to  interest 
on  the  money  expended,  and  take  the  purchaser's  note  for  the 
amount,  what  shall  be  the  iace  of  the  note,  and  what  its 
date  ?  ^^^      (  Face,  $2007.75. 

"*°  \  Date,  March  8,  185  G. 

G.   Thomas  Whiting, 
185'J,  To  IsKAEL  Palmer,     Dr. 

Jan.     lo     To    GO  bbls.  Flour,    O  $7.00,  . $420 

"     28.       "      90  bn.     Wheat,  "      1.50,  ...    .    135 
Mar  15.       "   300  bbls.  Flour,     "      6.00,  .  .     .  .  1800 
If  credit  of  3  months  be  given  to  each  item,  when  will  tho 
above  account  become  due  ?  Ans.   May  30. 

CASE    III. 

|gi?j©o  Wlicn  the  terms  of  credit  begin  at  different 
times,  and  the  account  has  both  a  debt  and  a  credit 
side. 

1..    Averanre  the  followina;  account. 


Dr. 


David  Wake. 


Cr. 


1858.     1 

June 

1 

t( 

16 

Oct. 

20 

To  Mdse 

«    Draft,  3  mo. . 
"    Cash, 


1858.     1 

400 

00 

1  July 

4 

800 

00 

Aug. 

20 

250 

00 

Sept. 

20 

By  Mdse. 
"   Cash. 


200 
150 
500 


00 
00 
00 


Focal 
date. 


Dr. 


oper.vtion. 


Cr. 


Due 

da. 

141 

31 

0 

Item.s. 

400 
800 
250 

1450 
850 

600 

Prod. 

Duo 

da. 

Items 

Prod. 

June    1 
Sept.  19 
Oct.   20 

56400 
24800 

July     4 
Aug.  20 
Sept.  20 

108 
61 
30 

200 
150 
500 

850 

21600 

9150 

15000 

45750 

Bal 

mces. 

81200 
45750 

35450 

35450  -i-  600  =z  59  da.,  average  term  of  interest. 
Oct.  20  —  59  da. 


Aug.  22,  balance  due. 


■V\rhat  is  Case  III  ?     Explain  operation. 


268 


EQUATION    OF   PAYMENTS. 


AxALYSls.  In  the  above  operation  ue  have  written  the  dates, 
showing  when  the  items  become  due  on  either  side  of  the  ac- 
count, adding  3  days'  grace  to  the  time  allowed  to  the  draft.  The 
latest  date,  Oct.  20,  is  assumed  as  the  focal  date  for  both  sides,  and 
the  two  columns  marked  da.  show  the  dili'ereuce  in  days  between 
each  date  and  the  focal  date.  The  products  are  obtained  as  in  the 
last  case,  and  a  balance  is  struck  between  tiie  items  charged  and  the 
products.  These  balances,  being  on  the  L)r.  side,  show  that  David 
Ware,  on  the  day  of  the  focal  date,  Oct.  20,  owes  i-GOO  with  interest 
on  $35450  for  1  day.  By  division,  this  interest  is  found  to  be  equal 
to  the  interest  on  .$600  for  59  days.  The  balance,  $600,  therefore, 
lias  been  due  59  daj's.  Reckoning  back  from  Oct.  12,  we  find  the 
date  when  the  balance  fell  due,  Aug.  22.     ]Ience  the  following 

Rule.  I.  Find  the  time  when  each  item  of  the  account  is 
due  ;  and  write  the  dates,  in  two  columns,  on  the  sides  of  the 
account  to  which  they  respectively  belong. 

II.  Use  either  the  earliest  or  the  latest  of  these  dates  as  the 
focal  date  fur  both  sides,  and  find  the  products  as  in  the  last 
case. 

III.  Divide  the  bcdance  of  the  products  by  the  balance  of  the 
account;  the  quotient  tvill  be  the  interval  of  time,  ivhich  must 
be  reckoned  from  the  focal  date  toward  the  other  dates  when 
both  balances  are  on  the  same  side  of  the  account,  but  from 
the  other  dates  when  the  balances  are  on  opposite  sides  of  the 
account. 

2.  "What  is  the  balance  of  the  following  account,  and  when 
is  it  due? 

John  Wilson. 

Dr.  Cr. 


1859. 

1859. 

Jan. 
Feb. 

it 

1 
4 

To  Mdse.  . 
"  Cash .  . 

448 
364 
232 

00 
00 
00 

Jan.i  20 

Feb.'  16 

"     l2o 

By  Am't  bvo't  forward 

"   1  Carriage 

"  Cash 

560 
264 
900 

00 
00 
00 

^^^^  ^  Balance,  $6.= 
'  \  Due  March  J 


$680. 
13. 


3.   If  the  followinj];  account  be  settled  by  giving  a  note,  what 
shall  be  the  face  of  the  note,  and  what  its  date? 


Give  analysis.     Rule. 


RATIO. 


2Gy 


Isaac  Foster. 


Dr 


Cr. 


18,5? 

• 

1858. 

Jan 

1 

To  Mdoc. 

on  o  ino. 

11.3 

80 

May 

11 

Ry  Cash 

11 

OO 

i< 

ii; 

ii              (. 

i»    ,3    u 

37 

48 

July 

112 

t*                it 

lo 

00 

June 

<j 

ii              ii 

(*    3    *' 

12 

2,5 

Oct. 

12 

((           if 

82 

00 

Aug. 

4 

((              <C 

((     9     ii 

66 

43 

Ans 


$154.07,  face  of  note. 
Mar.  26,  18j8,  duto. 


RATIO. 

SI5I .  Katio  is  the  comjiarison  with  each  other  of  two  num- 
bers of  tiie  same  kind.  It  is  of  two  Icinds  —  arithmetical  and 
geometrical. 

^5^.  Arithmetical  Eatio  is  the  difference  of  the  two 
numbers. 

Jl*!^.  Geometrical  E.atio  is  the  quotient  of  one  number 
divided  bj  the  other. 

S<3J:.  Wlien  Ave  use  the  Avord  ratio  alone,  it  implies  geo- 
metrical ratio,  and  is  exj^ressed  by  the  quotient  arising  from 
dividing  one  number  by  the  other.  Thus,  the  ratio  of  4  to  8 
is  2,  of  10  to  5  is  i,  &c. 

S*5S.    Ratio  is  indicated  in  two  ways. 

1st.  By  placing  two  points  between  the  numbers  compared, 
writing  the  divisor  before  and  the  dividend  after  the  points. 
Thus,  the-  ratio  of  5  to  7  is  written  5:7;  the  ratio  of  9  to 
4  is  written  9  :  4. 

2d.  In  the  form  of  a  fraction  ;  thus,  the  ratio  of  9  to  3  is  f  ; 
tlie  ratio  of  4  to  G  is  £. 

SIslH.    The  Terms  are  tlie  two  numbers  compared. 

S<>57.    The  Antecedent  is  flie  first  term. 

SslS.    The  Consequent  is  the  second  term. 

2I»1I>.  No  comparison  of  two  numbers  can  be  fully  ex- 
plained but  by  instituting  another  comparison  ;  thus,  the  com- 


NoTE.  It  is  thniiKht  best  to  omit  the  questions  at  the  bottom  of  tlie  pa^es.  in  the 
reiiiainiii;^  part  of  this  work,  leaving  the  teacher  to  use  such  as  may  be  deemed  ap- 
propnute.  ^ 


270  RATIO. 

parison  or  relation  of  4  to  8  cannot  be  fully  expressed  by  2, 
nor  of  8  to  4  by  h  If  the  question  were  asked,  what  relation 
4  bears  to  8,  or  8  to  4,  in  respect  to  magnitude,  the  answer  2, 
or  J-,  would  not  be  complete  nor  correct.  But  if  we  make 
i/ni/i/  the  standard  of  comparison,  and  use  it  as  one  of  the 
terms  in  illustrating  the  relation  of  the  two  numbers,  and 
say  that  the  ratio  or  relation  of  4  to  8  is  the  same  as  1  to  2, 
or  the  ratio  of  8  to  4  is  the  same  as  1  to  J-,  unify  in  both  cases 
being  the  standard  of  comparison,  then  the  whole  meaning  is 
conveyed. 

SIC5©.  A  Direct  Eatio  arises  from  dividing  the  consequent 
by  the  antecedent. 

261.  An  Inverse  or  Eeciprocal  Ratio  is  obtained  by  di- 
viding the  antecedent  by  the  consequent.  Thus,  the  direct 
ratio  of  5  to  15  is  -^-^-  =:  3 ;  and  the  inverse  ratio  of  5  to  15  is 
~^-  =  J-. 

3(|S|.  A  Simple  E,atio  consists  of  a  single  couplet ;  as 
3:12. 

S63.  A  Compound  Ratio  is  the  product  of  two  or  more 
simple  ratios.  Thus,  the  compound  ratio  formed  from  the 
simple  ratios  of  3  :  G  and  8:2isfXf  =  3X8:GX2  = 

JL2  —  1 

3^4.  In  comparing  numbers  with  each  other,  they  must 
be  of  tlie  same  lind,  and  of  the  same  denomination. 

S6«$.  The  ratio  of  two  fractions  is  obtained  by  dividing 
the  second  by  the  first ;  or  by  reducing  them  to  a  common  de- 
nominator, when  they  arc  to  each  other  as  their  numerators. 
Thus,  the  ratio  of  fV  =  "I  is  f  "i"  tV  =  t§  =  2,  which  is  the 
same  as  the  ratio  of  the  numerator  3  to  the  numerator  G  of 
the  equivalent  fractions  f'jj  and  -f^. 

Since  the  antecedent  is  a  divisor  and  the  consequent  a  divi- 
dend, any  change  in  cither  or  both  terms  will  be  governed  by 
the  general  principles  of  ilivision,  (^7.)  AV<'  liave  only  to 
substitute  the  terms  antecedent,  consequent,  and  i-atio,  for  divi- 
sor, dividend,  and  quotient,  and  these  principles  become 


RATIO.  271 


GENERAL   PllINCIPLES   OF   RATIO. 

Prin.  I.  MnUijyJy'mg  the  consequent  muUiplies  the  ratio; 
dividimj  the  consequent  divides  the  ratio. 

Piiix.  II.  Multiplying  the  antecedent  divides  the  ratio  ;  di- 
viding the  antecedent  multiplies  the  ratio. 

PiiiN.  III.  Multiphjing  or  dividing  both  antecedent  and  con- 
sequent by  the  same  number  does  not  alter  the  ratio. 

These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

A  change  in  the  consequent  produces  a  like  change  in  the 
ratio  ;  but  a  change  in  the  antecedent  produces  an  opposite 
change  in  the  ratio. 

JlCm.  Since  the  ratio  of  two  numbers  is  equal  to  the  con- 
sequent divided  by  the  antecedent,  it  follows,  that 

1.  The  antecedent  is  equal  to  the  consequent  divided  by 
the  ratio  ;  and  that, 

2.  The  consequent  is  equal  to  the  antecedent  multiplied  by 
tl\e  ratio. 

EXAMPLES    FOR   PRACTICE. 

1.    What  part  of  9  is  3? 
I  :=  i  ;  or,  9  :  3  as  1  :  J,  that  is,  9  has  the  same  ratio  to  3  that  1 
bi's  to  1^. 

2..  What  part  of  20  is  5  ?  Ans.   |. 

3.  What  part  of  36  is  4?  Ans.   ^. 

4.  What  part  of  7  is  49  ?  Ans.    7  times. 

5.  What  is  the  ratio  of  IG  to  88  ?  Ans.    5i 
G.   What  is  the  ratio  of  G  to  8^?  Ans.    \l. 

7.  What  is  the  ratio  of  64-  to  78?  Ans.    12. 

8.  What  is  the  ratio  of  16  to  66  ?  Ans.    4 >. 

9.  What  is  the  ratio  of  f  to  f  ?  Ans.    |. 

10.  What  is  the  ratio  of  I  to  jV?  ^"^^-    f- 

11.  What  is  the  ratio  of  3,1  to  1G|?  Ans.    5. 

12.  What  is  the  ratio  of  3  gal.  to  2  qt.  1  pt.  ?      Ans.   /j. 


272  PROPOETION. 

13.  What  is  the  ratio  of  6.3  s  to  8  s.  6  d.  7        Ans.    Iff 

14.  What  is  the  ratio  of  5.6  to  .56?  Ans.    y'^. 

15.  Wliat  is  the  ratio  of  19  lbs.  5  oz.  8  pwts.  to  25  lbs.  11 
oz.  4  pwts.  ?  Ans.   1^. 

16.  What  is  the  inverse  ratio  of  12  to  16?  Ans.   f. 

17.  What  is  the  inverse  ratio  of  f  to  |?  Ans.   ■^^. 

18.  What  is  the  inverse  ratio  of  5f  to  17^?         Ans.    \. 

19.  If  the  consequent  be  16  and  the  ratio  2f,  what  is  the 
antecedent?  Ans.   7. 

20.  If  the  antecedent  be  14.5  and  the  ratio  3,  what  is  the 
consequent?  Ans.    43.5. 

21.  If  the  consequent  be  J  and  the  ratio  f,  what  is  the  an- 
tecedent? Ans.    1|-. 

22.  If  the  antecedent  be  %  and  the  ratio  ^,  vrhat  is  the 
consequent?  Ans.   yV 

PROPORTION. 

S67.  Proportion  is  an  equality  of  ratios.  Thus,  the  ratios 
6  :  4  and  12  :  8,  each  being  equal  to  f,  form  a  proportion. 

3GS.    Proportion  is  indicated  in  two  ways. 

1st.  By  a  double  colon  placed  between  the  two  ratios  ;  thus, 
2  :  5  : :  4  :  10. 

2d.  By  the  sign  of  equality  placed  between  the  two  ratios ; 
thu.^s,  2:5  =  4:10. 

3@0.  Since  each  ratio  consists  of  two  terms,  every  pro- 
portion must  consist  of  at  least  four  terms. 

370.    The  Extremes  are  the  first  and  fourth  terms. 

S73.     The  Means  are  the  second  and  third  terms. 

3?s5.  Three  numbers  may  be  in  proportion  when  the  first 
is  to  the  second  as  the  second  is  to  the  third.  Thus,  the  num- 
bers 3,  9,  and  27  are  in  proportion  since  3  :  9  : :  9  :  27,  the 
ratio  of  each  couplet  being  3. 

In  such  a  proportion  the  second  term  is  said  to  be  a  7nean 
proportional  between  the  other  two. 

It7ti.  In  every  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means.  Thus,  in  the  proportion 
3  :  5  : :  6  :  10  we  have  3X10=5X6. 


SIMPLE   PROPORTION.  273 


«>>  > 


77^,  Four  numbers  that  are  iDroportional  in  the  direct 
order  are  pi'oportional  by  inversion,  and  also  by  alternation,  or 
by  inverting  tlie  means.  Thus,  tlie  proportion  2  :  3  :  :  6  :  9, 
by  inversion  becomes  3  :  2  : :  9  :  G,  and  by  alternation  2  :  G  : : 

t^73.  From  the  preceding  principles  and  illustrations,  it 
foliows  that,  any  three  terms  of"  a  jjroportion  being  given,  the 
fourth  mav  readily  be  found  by  the  followinfr 

lluLE.  I.  Divide  the  product  of  the  extremes  bi/  one  of  the 
means,  and  the  quotient  ivill  be  the  other  mean.     Or, 

1 1.  Divide  the  product  of  the  means  by  one  of  the  extremes, 
and  tlie  quotient  tcill  be  the  other  extreme. 

EXAMPLES    FOR    PRACTICE. 

Find  the  term  not  given  in  each  of  the  following  proportions. 

1.  48  :  20  :  :  (     )  :  50.  Ans.    120. 

2.  42  :  70  : :  3  :  (     ).  Ans.    5. 

3.  (     )  :  30  : :  20  :  100.  Ans.    6. 

4.  1  :  (    )  : :  7  :  84.  A)is.    12. 

5.  48  yd.  :(    )::  $67.25  :  $201.75.  Ans.    144  yd. 

6.  3  lb.  12  oz.  :  (     )  : :  $3.50  :  $10.50.     Ans.  11  lb.  4  oz. 

7.  (    )  :  $38.25  : :  8  bu.  2  pk.  :  76  bu.  2  pk.    Ans.  $4.25. 

8.  4i-  :  381  : :  (    )  :  7G^.  Ans.    8^. 

9.  (     )  :  12  ::  a:  If.  Ans.    7. 

10       -5-   •    ^       'i    •  •    '    •    2  /i)i<!       3 

^^'    iG  •  V     J  •  ■  ^  ■  b'  J±ns.    g. 

SliVIPLE   PROPORTION. 

157®.  Simple  Proportion  is  an  equality  of  two  simple 
ratios,  and  consists  of  four  terms,  any  three  of  which  being 
given,  the  fourth  may  readily  be  found. 

t^77.  Every  question  in  simple  proportion  involves  the 
principle  of  cause  and  effect. 

^7S.  CaAises  may  be  regarded  as  action,  of  whatever 
kind,  the  producer,  the  consumer,  men,  animals,  time,  distance, 
weight,  goods  bought  or  sold,  money  at  interest,  &c. 

SI'S.  Effects  may  be  regarded  as  whatever  is  accom" 
12* 


274  PROPORTION. 

plisbed  by  action  of  any  kind,  the  thing  produced  or  consumed, 
money  paid,  &c. 

el80.  Causes  and  effects  are  of  two  kinds  —  simple  and 
compound. 

S8i.  A  Simple  Cause,  or  Effect,  contains  but  one  element ; 
as  goods  purchased  or  sold,  and  the  money  paid  or  received 
for  them. 

S§3.  A  Compound  Cause,  or  Effect,  is  the  product  of  two 
or  more  elements ;  as  men  at  work  taken  in  connection  with 
time,  and  the  result  produced  by  them  taken  m  connection 
with  dimensions,  length  and  breadth,  &c. 

SI83.  Causes  and  effects  that  admit  of  computation,  that 
is,  involve  the  idea  of  quantity,  may  be  represented  by  num- 
bers, which  will  have  the  same  relation  to  each  other  as  the 
things  they  represent.  And  since  it  is  a  principle  of  philoso- 
phy that  like  causes  produce  lilce  effects,  and  that  effects  are 
always  ^V^  •proportion  to  their  causes,  we  have  the  following 
proportions : 

1st  Cause  :  2d  Cause  :  :  1st  Effect  :  2d  Effect. 
Or,  1st  Effect  :  2d  Effect  :  :  1st  Cause  :  2d  Cause;  | 

in  which  the  two  causes,  or  the  two  effects  forming  one  coup- 
let, must  be  like  numbers,  and  of  the  same  denomination. 

Considering  all  the  terms  of  the  proportion  as  abstract  num- 
lers,  we  may  say  that 

1st  Cause  :  1st  Effect  :  :  2d  Cause  :  2d  Effect, 
which  will  produce  the  same  numerical  result. 

But  as  ratio  is  the  result  of  comparing  two  numbers  or 
things  of  the  same  kind  (364),  the  first  form  is  regarded  as 
the  most  natural  and  philosophical. 

384:.  Simple  causes  and  simjile  effects  give  rise  to  simple 
ratios ;  compound  causes  and  compound  effects  to  compound 
ratios. 

385.     1.  If  5  tons  of  coal  cost  $30,  what  will  3  tons  cost  ? 

Note.  The  required  term  will  be  denoted  by  a  (  ),  and  designated 
"  blank." 


1 


SIMPLE  pr.oroiiTiON. 


275 


STATEMENT. 

tons.        tons.  $  $ 

5     :     3    :  :    30    :    (     ) 

1st  cause.  2(1  cause.     1st  effect.    2a  effect. 
OPERATION. 

5  X  (     )  =  3  X  30 

(     )  z= =^  S18,  Ans. 


Analysis.  In  tliis 
example  an  ejj'cd  is 
required,  and  5  tons 
must  have  the  same 
ratio  to  3  tons,  as 
$30,  the  cost  of  5 
tons,  to  (blank)  dol- 
lars, the  cost  of  3 
tons. 


STATEMENT. 

bar. 

bar.                 S 

% 

15      : 

(     )  :  :    90     : 

30 

st  cause. 

2(1  cause.      1st  effect. 
OPERATION. 

2a  effect 

00 

^0 

( 

) 

W> 

( 

)  = 

-  5  bar.,  Ans. 

Since  the  product  of  the  extremes  is  equal  to  the  product  of  the 
means  (3^3),  and  the  product  of  the  means  divided  by  one  of  the 
extremes  will  give  the  other;  (blank)  dollars  wiU  be  equal  to  the 
product  of  3  X  30  divided  by  5,  which  is  $18,  Aas. 

2.    If  15  barrels  of  flour  cost  $90,  how  many  barrels  can  be 

bought  for  $30  ? 

Analysis.  In  this  ex- 
ample a  cause  is  required, 
and  the  statement  may  be 
read  thus :  If  15  barrels 
cost  $90,  how  many  or 
(blank)  barrels  will  cost 
$30  ?  The  product  of  the 
extremes,  30  X  15,  di- 
vided by  the  given  mean, 
90,  will  give  the  required 

term,  5,  as  shown  in  the  operation.    Hence  we  deduce  the  following 

Rule.  I.  Arrange  the  terms  in  the  statement  so  that  the 
causes  shall  compose  one  coi/plef,  and  the  effects  the  other,  put- 
ting (     )  in  the  place  of  the  required  term. 

II.    If  the  required  term  he  an  extreme,  divide  the  -product 

of  the  means  hy  the  given  extreme  ;  if  the  required  term  he  a 

mean,  divide  the  product  of  the  extremes  hy  the  given  mean. 

Notes.  1.  If  the  terms  of  any  couplet  be  of  different  denominations, 
they  must  be  reduced  to  the  same  unit  value. 

2.  If  the  odd  term  be  a  compound  ninnbcr,  it  must  be  reduced  to  its 
lowest  unit. 

3.  If  the  divisor  and  dividend  contain  one  or  more  factors  common 
to  both,  they  should  be  canceled.  If  any  of  the  tciTns  of  a  proportion 
contain  mixed  numbers,  they  should  first  be  changed  to  improper  frac- 
tions, or  the  fractional  part  to  a  decimal. 

4.  "NVhen  the  vertical  line  is  used,  the  divisor  and  the  required  terra 
are  written  on  the  left,  and  the  terms  of  the  dividend  on  the  right. 


276  PEOPORTION. 


«T9 


We  will  now  give  another  method  of  solving  ques- 
tions in  simple  proportion,  without  making  the  statement,  and 
which  may  be  used,  by  those  who  prefer  it,  to  the  one  ah-eady 
given.     >Ye  v/ill  term  it  the 

Second  Method. 

Eveiy  question  which  properly  belongs  to  simple  propor- 
tion must  contain  four  numbers,  at  least  three  of  which  must 
be  given  (IIT6).  Of  the  three  given  numbers,  one  must 
always  be  of  the  same  denomination  as  the  required  number. 
The  remaining  two  will  be  like  numbers,  and  bear  the  same 
relation  to  each  other  that  the  third  does  to  the  required  num- 
ber ;  in  other  words,  the  ratio  of  the  third  to  the  required 
number  will  be  the  same  as  the  ratio  of  the  other  tv/c  num^ 
bers. 

Regarding  the  third  or  odd  term  as  the  antecedent  of  the  sec- 
ond couplet  of  a  proportion,  we  find  the  consequent  or  required 
term  by  multiplying  the  antecedent  by  the  ratio  (SGd). 

By  comparing  the  two  like  numbers,  in  any  given  question, 
\\'ith  the  third,  we  may  readily  determine  whether  the  answer, 
or  required  term,  will  be  greater  or  less  than  the  third  term  ; 
if  greater,  then  the  ratio  will  be  greater  than  1,  and  the  two 
like  numbers  may  be  arranged  in  the  form  of  an  improjier 
fraction  as  a  multiplier ;  if  the  answ^er,  or  required  term,  is  to 
be  less  than  the  third  term,  then  the  ratio  wnll  be  less  than  1, 
and  the  two  like  numbers  may  be  arranged  in  the  form  of  a 
proper  fraction,  as  a  multiplier. 

1.    If  4  cords  of  wood  cost  $12,  what  wall  20  cords  cost? 

OPERATION.  AnATA'SIS.      It  will 

t^^  X  20  be  readilv  seen  in  tliis 

12  X  -^-j  written =  $60.         examiilc,"that  4  cords 

^  and  20  cords  arc  the 

like  terms,  and  that 
$12  is  the  third  term,  and  of  the  same  denomination  as  the  answer 
or  required  term. 

If  4  cords  cost  $12,  will  20  cords  cost  more,  or  less,  than  4  cords? 
evidently  more :  then  the  answer  or  required  term  will  be  greater 


SIMPLE   PROPORTION.  277 

than  the  third  term,  and  the  ratio  greater  than  1.  The  ratio  of  4 
cords  to  20  cords  is  ^O-,  or  5 ;  hence  the  ratio  of  $12  to  the  answer 
must  be  o,  and  the  answer  will  be  ^^  or  5  times  $12,  which  is  $60. 

2.   If  12  yards  of  cloth  cost  $48,  what  will  4  yards  cost  ? 

OPERATION.  Analysis.     In  this   example  we 

4-8  X  1*2  =  $16,  Ans.         see  that  12  yards  and  4  yards  are 

the   like   terms   and  $48  the  third 
term,  and  of  the  same  denomination  as  the  required  answer. 

If  12  yards  cost  $48,  will  4  yards  cost  more  or  less  than  12  yards? 
less:  then  the  ratio  will  be  less  than  1,  and  the  multiplier  a  proper 
fraction.  The  ratio  of  12  yards  to  4  yards  is  -^% ;  hence  the  ratio  of 
$48  to  the  answer  is  ■^-^,  and  the  answer  will  be\%  times  $48,  which 
is  $16.     Hence  the  following 

Rule.  I.  Jiith  the  (wo  given  manhers,  which  are  of  the 
same  name  or  kind,  form  a  ratio  greater  or  less  than  1,  accord- 
ing as  the  answer  is  to  be  greater  or  less  than  the  third  given 
number. 

II.  MaUlpty  the  third  number  by  this  ratio,  and  the  product 
loill  be  the  required  number  or  answer. 

Note.  1.  !Mixed  numbers  should  first  be  reduced  to  improper  frac- 
tions, and  the  ratio  of  the  fractions  foimd  according  to  (365). 

2.  Reductions  and  cancellation  may  be  applied  as  in  the  first  method. 

The  following  examples  may  be  solved  by  either  of  the 
foregoiug  methods. 

EXAMPLES    FOB    PRACTICE. 

1.  If  48  cords  of  wood  cost  $120,  how  much  will  20  cords 
cost?  Ans.    $50. 

2.  If  6  bushels  of  corn  cost  $4.7^'>,  how  much  will  75  bush- 
els cost  ?  Ans.    $59.37^-. 

3.  If  8  yards  of  cloth  cost  $3J-,  how  many  yards  can  be 
bought  for  $50  ?  Ans.    114f  yds. 

4.  If  12  horses  consume  42  bushels  of  oats  in  3  weeks,  how 
many  bushels  will  20  horses  consume  in  the  same  time  ? 

5.  If  7  pounds  of  sugar  cost  75  cents,  how  many  pounds 
can  be  bought  for  $9  ?  Ans.    84  lbs. 

6.  What  will  11  lb.  4  oz.  of  tea  cost,  if  3  lb.  12  oz.  cost 
$3.50?  Ans.   $10.50. 


278  SIMPLE   PROPORTION. 

7.  If  a  staff  3  ft.  8  in.  long  cast  a  shadow  1  ft.  6  in.,  ■\vliat 
is  the  heiglit  of  a  steeple  that  casts  a  shadow  75  feet  at  the 
same  time  ?  Ans.    183  ft.  4  in. 

8.  At  $2.75  for  14  pounds  of  sugar,  what  will  be  the  cost 
of  100  pounds?  ■  A71S.    $19.64f. 

9.  How  many  bushels  of  wheat  can  be  bought  for  $5 LOG, 
if  12  bushels  can  be  bought  for  $13.32  ? 

10.  What  will  be  the  cost  of  28  J^  gallons  of  molasses,  if  15 
hogsheads  cost  $23G.25  ?  Aus.    §7.12i-. 

11.  If  7  barrels  of  flour  are  sufficient  for  a  family  6  mouths, 
how  many  barrels  will  they  require  for  11  months? 

12.  At  the  rate  of  9  yards  for  £5  12  s.,  how  many  yards  of 
cloth  can  be  bought  for  £44  16  s.  ?  Ans.    72  vds. 

13.  An  insolvent  debtor  fails  for  $75 GO,  of  which  he  is 
able  to  pay  only  $3100 ;  how  much  will  A  receive,  whose 
claim  is  $75G  ?  Ans.    $310. 

14.  If  2  pounds  of  sugar  cost  25  cents,  and  8  pounds  of 
sugar  are  worth  5  pounds  of  coffee,  what  will  100  pounds  of 
coffee  cost  ?  Ans.    $20. 

15.  If  tlie  moon  move  13°  10'  35''  in  1  day,  in  what  time 
will  it  perform  one  revolution  ? 

IG.  If  8^  bushels  of  corn  cost  $4.20,  what  will  be  the  cost 
of  13^  bushels  at  the  same  rate  ?  Ans.    $6.48. 

17.  If  IJ  yards  of  cotton  cloth  cost  G]-  pence,  how  many 
yards  can  be  bought  for  £10  6  s.  8  d.  ?  Ans.  694|yds. 

18.  If  124-  cwt.  of  iron  cost  $42^,  how  much  will  4S^  cwt. 
cost?  Ans.    $163,50  4". 

19.  What  quantity  of  tobacco  can  be  bought  for  $317.23, 
if  8|  lbs.  cost  $lf  ?  Ans.    15  cwt.  22.7-f-  lbs. 

20.  If  \r^l  bushels  of  clover  seed  cost  $156],  how  inm-h 
can  be  bought  for  $95.75  ?  Ans.    9  bu.  2  pk.  2§  <it. 

21.  If  I  of  a  barrel  of  cider  cost  $fS,  how  mucli  will  |  of 
a  barrel  cost?  An^.    S  ''  . 

22.  If  a  piece  of  land  of  a  certain  length,  and  4  roils  in 
bieadlli,  contain  |  of  an  acre,  Low  nuich  would  tlicrc  bo  if  it 
were  11  f  rods  wide  ?  Ans.    2  A.  28  rods. 

23.  If    13    cwt.  of  iron  cost  $42|,  what  will    12   cwt.  cost  ? 


SIMrLE   PROPORTION. 


279 


24.  A  grocer  has  a  false  balance,  by  wliicli  1  poiiiul  -will 
■vveigii  but  12  oz.  j  what  is  the  real  value  of  a  barrel  of  sugar 
that  he  sells  for  $28  ?  Ans.    $21. 

25.  A  butcher  in  selliug  meat  sells  14|^  oz.  fur  a  pound  ; 
how  mufh  (Iocs  he  cheat  a  customer,  who  buj's  of  liiui  to  the 
amount  of  $oO?  Aus.    $2A(j -\- . 

2G.  If  a  man  clear  $750  by  bis  business  iu  1  yr.  G  mo.,  how 
much  would  he  gain  in  3  yi\  9  mo.  at  the  same  rate  ? 

27.  If  a  certain  business  yield  $350  net  profits  in  10  mo., 
in  what  time  would  the  same  business  yield  $1050  profits  ? 

28.  B  and  C  have  each  a  farm ;  B's  farm  is  worth  $25 
an  acre,  and  C's  $30^  ;  if  in  trading  B  values  his  land  at  $28 
an  acre,  wdiat  value  should  C  put  upon  his  ?     Ans.    $34.10. 

29.  If  I  borrow  $500,  and  keep  it  1  yr.  4  mo.,  for  how  long 
a  time  should  I  lend  $240  as  an  eauivalent  for  the  favor  ? 

Ans.    2yr.  9  mo.  10  da. 


COMPOUND   PROPORTION. 

S8T,  Compound  Proportion  embraces  that  class  of  ques- 
tions in  which  the  causes,  or  the  effects,  or  both,  are  compound. 

The  required  term  may  be  a  cause,  or  a  single  element  of  a 
cause  ;   or  it  may  be  an  effect,  or  a  single  element  of  an  elfect. 

1.  If  IG  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ? 


1st  cause. 


STATEMENT. 

2d  cause.  1st  effect.      2d  effect 


(    5( 


16    .        f     5 

50     *        1    90 
IG  X  50     :    5  X  90 

OPERATION. 

fj  X  00^  X  3'i^S^ 


128 


Or, 


(     ) 
(     ) 


;    128 

Analysts.     In  this  ex- 
ample the  required  term 

72    bu.  i'^  the  second  effect;  and 

i{i  X  00  the  question  may  be  read, 

If   16  horses   in  50  days 
consume  128  bushels  of  oats,  5  horses  in  90  days  mIII  consume 
how  many,  or  (blank)  busliels  ? 
XoTE.     These  questions  arc  most  readily  performed  by  cancellation. 


280  PROPORTION. 

2.  If  $480  gain  $84  interest  in  30  months,  wliat  sum  -will 
gain  S21  in  15  niontlis  ? 

STATEMENT. 
1st  cause.         2d  cause.         Ist  efifect.     2d  effect. 

OPERATiox.  Analysis.    The  re- 

4^(i^-'^  X  ^(i~  X  ^i  quired  term  in  tliis  cx- 

— z=  $2-10,  Ajis.         ample  is  an  element  of 

$4  X  2l$  the  second  cause;  and 

the  question  may  he 
read,  If  $480  in  30  months  gain  $84,  -nhat  principal  in  1-j  months 
will  gain  $21  ? 

3.  If  7  men  dig  a  ditch  GO  feet  h)ng,  8  feet  wide,  and  G  feet 
deep,  in  12  days,  what  length  of  ditch  can  21  men  dig  in  2§ 
days,  if  it  be  3  feet  wide  and  8  feet  deep  ? 

STATEMENT. 

21  CGO       (    (     ) 


UA 


2^  :  :    ■<     S  :    ^        o 


G       (       8 
Or,  7  X  12   :  21  X  §  :  :   GO  X  8  X  G  :    (      )  X  3  X  8 

OPERATION.  ■      Analysis.    In 

^t  X     8  X  00'^  X  $  X  iS^  this  example  the 

— — —  zn  80  ft.,  Ans.     required  term  is 

//  X  J'2  X    3     X  ^  X  3  _  the  length  of  the 

Or        3       8  ditch,  and  is  an  element  of  the 

second  effect.  The  question, 
as  stated,  will  read  thus :  if  7 
men,  in  12  days,  dig  a  ditch  GO 
feet  long,  8  feet  wide,  and  G 
feet  deep.  21  men,  in  2f  days, 
will  dig  a  ditch  how  many,  cr 
(hlank)  feet  long,  3  feet  wide, 
(      )  =  80  ft.,  Ans.  and  8  feet  deep? 

Hence  we  have  the  following 

Rule.  I.  Of  (lie  (jiven  terms,  select  those  tohich  constitute 
the  causes,  and  those  which  constitute  the  effects,  and  ai-range 
them  in  couplets,  putting  (     )  in  place  of  the  required  term. 


B 

8 

71 

^t 

X^ 

00' 

$ 

$ 

^ 

0^ 

(     ) 

COMPOUND   PROPORTION.  281 

II.  Tlien,  if  the  hlanh  term  (  )  occur  in  either  of  the  ex- 
tremes, make  the  jiroduct  of  the  means  a  dividend,  and  the 
product  of  the  extremes  a  divisor  ;  hut  if  the  hlanh  term  occur 
in  either  mean,  make  tJie  product  of  the  extremes  a  dividend, 
and  the  product  of  the  means  a  divisor. 

Notes.  1.  The  causes  must  be  exactly  alike  in  the  munber  and  kind 
of  their  terms  ;  the  same  is  true  of  the  effects. 

2.  The  same  preparation  of  the  terms  by  reduction  is  to  be  observed 
as  in  simple  proportion. 

SI88.  We  will  now  solve  an  example  according  to  the 
Second  Method  given  in  Simple  Proportion. 

1.  If  18  men  can  build  42  rods  of  wall  in  16  days,  how 
many  men  can  build  28  rods  in  8  days .'' 

OPERATION.  Analysis.     We  see  in 

^$^       ^0'^  this  example  that  all  the 

^B    X  ——   X  — -  =  21  men.  terms  appear  in  couplets, 

^•^  "  except   one,  Avhich   is   IS 

men,  and  that  is  of  the  same  kind  as  the  required  answer. 

Since  compound  proportion  is  made  up  of  two  or  more  simple 
proportions,  if  this  third  or  odd  term  be  multiplied  by  the  compound 
ratio,  or  by  the  simple  ratio  of  each  couplet  successively,  the  prod- 
uct will  be  the  required  term. 

By  comparing  the  terms  of  each  couplet  with  the  third  term  Me 
may  readily  determine  whether  the  answer,  or  term  sought,  will  be 
greater  or  less  than  the  third  term  ;  if  greater,  then  the  ratio  will  be 
greater  than  1,  and  the  multii^lier  an  improper  fraction  ;  if  less,  the 
ratio  will  be  less  than  1,  and  the  multiplier  a  proper  fraction. 

First  we  will  compare  the  terms  composing  the  first  couplet,  42 
rods  and  28  rods,  with  the  third  term,  18  men.  If  42  rods  require 
18  men,  how  many  men  will  28  rods  require?  less  men;  hence  the 
ratio  is  less  than  1,  and  the  multiplier  a  proper  fraction,  || ;  next, 
if  16  days  require  18  men,  how  many  men  will  8  days  require? 
more  men  ;  hence  the  ratio  is  greater  than  1,  and  the  multiplier  an 
improper  fraction,  L<i.  Regarding  the  third  term  as  the  antecedent 
of  a  couplet,  the  consequent  being  the  term  sought,  if  Me  multiply 
this  third  term  by  the  simple  ratios,  or  by  their  product,  m'c  shall 
have  the  required  term  or  ansM'er,  thus :  18  X  f|  X  ^*^  =:  24,  as 
shown  in  the  operation. 

2.  5  compositors,  in  16  days,  of  14  hours  each,  can  compose 
20  sheets  of  24  pages  in  each  sheet,  50  lines  in  a  page,  and 


282  COMPOUND   PROPORTION. 

40  letters  in  a  line ;  in  how  many  days,  of  7  hours  each,  will 
10  compositors  compose  a  volume  to  be  printed  in  the  same 
letter,  containing  40  sheets,  IG  pages  in  a  sheet,  GO  lines  in  a 
page,  and  50  letters  in  a  line  ?  A7is.  32  days. 

OPERATIOX, 
davs.     comp.     hours,    sheets,    pages.     lines,     letters. 

lOX  T^o  X  V-  X  la  X  if  X  f  a  X  la  =  32  days. 

BY  CANCELLATION.  ANALYSIS.     The  required  term  or  an- 

16  swer  is  to  be  in  days  ;  and  we  see  that 

v/i       rt;  all  the  terms  appear  in  pairs  or  couplets, 

,      ,  ,  except  the  16  days,  which  is  of  the  same 

^        ^  kind  as  the  answer  sought. 

^0    40  We  will  proceed  to  compare  the  terms 

^4    ■3^0^  of  each  couplet  with  the  IG  days.     Fhst, 

/^ri    ripi  if  5  compositors  require   16    days,  how 

many  days  wdl  10  compositors  require  ? 
^"    ^^  less  days ;    hence  the   multiplier  is  the 

32  days,  A7is.  proper  fraction  ^\,  and  we  have  16  X  iV 
Next,  if  14  hours  a  day  require  16  days, 
how  many  days  will  7  hours  a  day  require  ?  more  days  ;  hence  the 
muItipHer  is  the  improper  fraction  -y-,  and  we  have  16  X  yo  X  V"- 
Next,  if  20  sheets  require  16  daj's,  how  many  days  will  40  sheets 
require  ?  more  days ;  hence  the  multiplier  is  the  im])roper  fraction 
•|^,  and  we  have  16  X  \o  X  V"  X  ff •  Pursuing  the  same  method 
with  the  other  couplets,  we  obtain  the  result  as  shown  m  the  opera- 
tion.    Hence  we  have  the  following 

Rule.  I.  Of  the  terms  composing  each  couplet  .form  a 
ratio  greater  or  less  than  1,  in  the  same  manner  as  if  the 
answer  depended  on  those  two  and  the  third  or  odd  term. 

II.  Midliply  the  third  or  odd  term  hy  these  ratios  successivehj, 
and  the  product  will  he  the  answer  sought. 

Note.  By  the  odd  term  is  meant  the  one  that  is  of  the  same  kind 
as  the  answer. 

The  following  examples  ma}'-  be  solved  by  either  of  tlie 
given  methods. 

EXAMPLES    FOR    TUACTICE. 

1.  If  IG  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ? 


COMrOUND   PROPORTION.  283 

2.  If  a  man  (ravel  120  miles  in  3  clays  when  the  days  are 
12  hours  long,  in  how  many  clays  of  10  hours  each  will  he 
require  to  travel  3 GO  miles'?  Ans.  lOA  days. 

3.  If  G  laborers  dig  a  ditch  34  yards  long  in  10  days,  how 
many  yards  can  20  laborers  dig  in  15  days?     Ans.  170  }ds. 

4.  If  450  tiles,  each  12  inches  squai-e,  will  pave  a  cellar, 
how  many  tiles  that  are  9  inches  long  and  8  inches  wide  will 
pave  the  same  ?  Ans.    900. 

5.  If  it  require  1200  yards  of  cloth  |-  wide  to  clothe  500 
men,  how  many  yards  which  is  ^  wide  will  it  take  to  clothe 
OGOmen?  Ans.    3291f  yds. 

6.  If  8  men  Avill  mow  36  acres  of  grass  in  9  days,  of  9 
hours  each  day,  how  many  men  will  be  required  to  mow  48 
acres  in  12  days,  working  12  hours  each  day?       Ans.  6  men. 

7.  If  4  men,  in  24  days,  mow  G|  acres  of  grass  by  work- 
ing 8;^  hours  a  day,  how  many  acres  will  15  men  mow  in  3|- 
days  by  working  9  hours  a  day?  Ans.    4.0 {^  acres. 

8.  If,  by  traveling  6  hours  a  day  at  the  rate  of  41  miles 
an  hour,  a  man  perform  a  journey  of  540  miles  in  20  days,  in 
how  many  days,  traveling  9  hours  a  day  at  the  rate  of  4g- 
miles  an  hour,  will  he  travel  GOO  miles  ?        Ans.    14f  days.  . 

9.  If  21  yards  of  cloth  If  yards  wide  cost  $3.37^,  what 
cost  3G1  yards,  1^  yards  wide?  Ans.    $52.79  -\-. 

10.  If  5  men  reap  52.2  acres  in  G  days,  how  many  men 
will  reap  417.6  acres  in  12  days  ?  Ans.    20  men. 

11.  If  G  men  dig  a  cellar  22.5  feet  long,  17.3  feet  wide,  and 
10.25  feet  deep,  in  2.5  days,  of  12.3  hours,  in  how  many  days, 
of  8.2  hours,  will  9  men  take  to  dig  another,  measuring  45 
feet  long,  34. G  wide,  and  12.3  deep?  A)is.    12  days. 

12.  If  54  men  can  build  a  fort  in  24i  days,  working  124 
hours  each  day,  in  how  many  days  will  75  men  do  the  same, 
when  they  work  but  lOi  hours  each  day  ?        Ans.  21  days. 

13.  If  24  men  dig  a  trench  33|-  yards  long,  5-|  wide,  and 
31  deep,  in  189  days,  working  14  hours  each  day,  how  many 
hours  per  day  must  217  men  work,  to  dig  a  trench  23^  yards 
I"ng,  3g-  wide,  and  2^  deep,  in  54  days  ?       Ans.    IG  hours. 


284  PARTNERSHIP. 

PARTXERSmP. 

389.  Partnership  is  a  relation  established  between  two 
or  more  persons  in  trade,  by  wliicli  they  agree  to  share  the 
profits  and  losses  of  business. 

II1>0.    The  Partners  are  the  individuals  thus  associated. 

?5©1.  Capital,  or  Stock,  is  the  money  or  property  invested 
in  trade. 

S92.    A  Dividend  is  the  profit  to  be  divided. 

*^93.    An  Assessment  is  a  tax  to  meet  losses  sustained. 


«s 


CASE    I. 

394.  To  find  each  partner's  share  of  the  profit  or 
loss,  when  their  capital  is  employed  for  equal  periods  of 
time. 

1.  A  and  B  engage  in  trade;  A  furnishes  $300,  and  B 
$400  of  the  capital;  they  gain  $182;  what  is  each  one's 
share  of  the  profit  ? 

Analysis.  Since 
the  whole  capital 
employed  is  $300 
+  $400  =  $700,  it 
is  evident  that  A 
furnishes  i-^^  =z  I 
of  the  capita],  and 

$182  X  i  =     $78,  A's  share  of  the  gain.  ^    fw  =  t   of  the 

$182Xf  =  $104,B-s    ..     "      .=  'T'-'^^-  :^"f^'"^! 

^  ^        ^        '  eacn  man  s  share  of 

the  profit  or  loss  will  have  the  same  ratio  to  the  whole  profit  or 
loss  that  his  share  of  the  stock  has  to  the  whole  stock,  A  will  have 
■2-  of  the  entire  profit,  and  B  i,  as  shown  in  the  operation. 

We  may  also  regard  the  whole  capital  as  the  first  cause, 
and  each  man's  share  of  the  capital  as  the  second  cause,  the 
whole  profit  or  loss  as  the  frst  effect,  and  each  man's  share  of  the 
profit  or  loss  as  the  second  effect,  and  solve  by  proportion  thus ; 


OPERATION. 

.  $300 

$400 

$700, 

whole 

stock. 

300 
Too 

=  h 

A's 

share  of  the  stock 

Too 

=  h 

B-8 

<(           <(             a 

PARTNERSHIP. 

let  cause. 

2d  cause.            1st  effi  ct. 

2d  effect. 

$700 

:     $300     :  :     $182 

:     (     ) 

$700 

:     $400     :  :     $182 

:     (     ) 

/100 

^00' 

/^00 

^00^ 

(      ) 

A$2~^ 

(     ) 

l^^'^s 

285 


(      )  —  ^lO**  B's  profit. 


(      )  —  ^  '  8,    A'8  profit. 

Hence  we  have  the  folio  win  a; 

Rule.  Multiply  the  whole  profit  or  loss  hy  the  ratio  of  the 
lohole  capital  to  each  mail's  share  of  the  capital.     Or, 

21ie  whole  capital  is  to  each  man's  share  of  the  capital  as  the 
whole  profit  or  loss  is  to  each  mans  share  of  the  profit  or  loss. 

.2  Three  men  trade  in  company;  A  furnishes  $8000,  B 
$12000,  and  C  20000  of  the  capital;  their  gain  is  $1G80; 
Avhat  is  each  man's  share  ? 

Ans.   A's  $33G  ;  B's  $504 ;  C's  $840. 

3.  Three  persons  purchased  a  house  for  $2800,  of  Avhich  A 
paid  S1200,  B  $1000,  and  C  $600 ;  they  rented  it  for  $224 
a  year ;  how  much  of  the  rent  should  each  receive  ? 

4.  A  man  failed  in  business  for  $20000,  and  his  available 
means  amounted  to  only  $13654 ;  how  much  will  two  of  his 
creditors  respectively  receive,  to  one  of  whom  he  owes  $3060, 
and  to  the  other  $1530  ?  Ans.   $2089.062  ;  $1044.531. 

5.  Four  men  hired  a  coach  for  $13,  to  convey  them  to 
their  respective  homes,  which  were  at  distances  from  the  place 
of  starting  as  follows:  A's  16  miles,  B's  24  miles,  C's  28 
miles,  and  D's  36  miles;    what  ought  each  to  pay? 

. ,     ^  A  $2.      C.  $3.50. 
"^*  1 B  $3.      D  $4.50. 

6.  A  captain,  mate,  and  12  sailors  took  a  prize  of  $2240, 
of  which  the  captain  took  14  shares,  the  mate  6  shares,  and 
each  sailor  1  share  ;  how  much  did  each  receive  ? 

7.  A  cargo  of  corn,  valued  at  $3475.60,  was  entirely  lost ; 
^  of  it  belonged  to  A,  ^  of  it  to  B,  and  the  remainder  to  C  ; 
how  much  was  the  loss  of  each,  there  being  an  insurance  of 
$2512?     ^«s.  $120.45,  A's.     $240.90,  B's.     $602.25,  C's. 


286  PARTNERSHIP. 

8.  Three  persons  engaged  in  the  lumber  trade  ;  two  of  the 
persons  furnished  the  capital,  and  the  third  managed  the  busi- 
ness; they  gained  $2571. 2-i,  of  which  C  received  $6  as  often 
as  D  $4,  and  E  had  -i  as  much  as  the  otlier  two  for  taking  care 
of  the  business  ;  how  mueli  was  eacli  one's  share  of  the  gain  ? 

Ans.  $1285.62,  C's.     $857.08,  D's.     $428.54,  E's. 

9.  Four   persons    engage    in  the    coal   trade ;    D  puts  in 

$3042  capital ;  they  gain  $7500,  of  which  A  takes  $2000,  B 

$2800.75,  and  C  $1685.25  ;   how  much  capital  did  A,  B,  and 

C  put  in,  and  how  much  is  D's  share  of  the  gain  ? 

.       ( A,  $6000.  C,  $5055.75. 

^"^-  X  B,  $8402.25.     D's  gain,  $1014. 

CASE    II. 

09«]».  To  find  each  partner's  share  of  the  profit  or 
loss  when  their  capital  is  employed  for  unequal  periods 
of  time. 

It  is  evident  that  the  respective  shares  of  profit  and  loss  will 
depend  upon  two  conditions,  viz. :  the  amount  of  capital  in- 
vested by  each,  and  the  time  it  is  employed. 

1.  Two  persons  form  a  partnership;  A  puts  in  $450  for 
7  months,  and  B  $300  for  9  months;  they  lose  $156 ;  how 
much  is  each  man's  share  of  the  loss  ? 

oPERATiox.  Analysis.     The 

$150  X  7  =  $3150,  As  oapit.-ii  for  1  mo.  "se  of  $4 JO  Capital 

$300  X  9  =  S2700, 15-3     «      «     «  for  7  months  is  the 

same  as  the  use  of 
'     "  7    times    $4,30,    or 

apitai  $3150  for  1  month; 

„    ■  and  of  $300  for  9 

months  is  the  same 
as  the  use  of  0  times 
$300,  or  $2700  for 
1  month.     The  en- 
tire capital  for  1  month  is  equivalent  to  $3150 -|- $2700  r=  .$58  JO. 
If  the  loss,  $150,  l)e  divided  lietwcen  the  two  partners,  according  to 
Case  I,  the  results  will  be  the  loss  of  each  as  shown  in  the  operation. 


$5850,  en;iio  « 

31  5.0   

5850   

Y^j,  A's  share  of  the  ent 

2  7  00   _ 
5if  50   

T3'  ^'^      "        " 

$156  X 

YjT  =  $84,  A's  loss. 

$156  X 

T^;y=$72,    D's      " 

PARTNERSHIP.  287 

Examples    of  this    kind   may  also  be  solved  by  proportion  as  in 
Case  I,  the  causes  being  compounded  of  capital  and  iijiie;  thus, 

$5850  :  $3150  :  :  $156  :  (  ) 
$5850  :  $2700  :  :  $15G  :  (  ) 


(   ) 


(       )  =  $8i,  A'a  loss.  (       )  =  $12,  B's  loss. 

Ilcnce  the  following 

Rule.  Multiply  each  man's  capital  hy  the  time  it  is  em- 
piloyed  ill  trade,  and  add  the  jn'oducts.  Then  multijily  iJ'C 
entire  profit  or  loss  by  the  ratio  of  the  sum  of  the  products  to 
each  product,  and  the  results  will  be  the  respective  shares  of 
profit  or  loss  of  each  piartner.     Or, 

Multiply  each  man's  capital  by  the  time  it  is  employed  in 
trade,  and  regard  each  product  as  his  capital,  and  the  sum  of 
the  products  as  the  entire  capital,  and  solve  by  proportion,  as  in 
Case  I. 

EXAJIPLES    FOR    rEACTICE. 

2.  Three  persons  traded  togetlier;  B  put  in  $250  for  6 
months,  C  $275  for  8  months,  and  D  $450  for  4  months ; 
they  gained  $825  ;  how  much  was  each  man's  share  of  the, 
gain  ? 

3.  Two  merchants  formed  a  partnership  for  1 8  months.  A  at 
lir,--t  put  in  $1000,  and  at  the  end  of  8  months  he  put  in  $000 
more ;  B  at  first  put  in  $1500,  but  at  the  end  of  4  months  he 
drew  out  $300 ;  at  the  expiration  of  the  time  they  found  that 
they  liad  gained  $1394.64;  how  much  was  each  man's  share 
of  the  gain  ?  Ans.    A's  $715.20  ;  B's  $679.44. 

4.  Three  men  took  a  field  of  grain  to  harvest  and  thresh 
for  i  of  the  crop ;  A  furnished  4  hands  5  days,  B  3  lianda 
6  days,  and  C  6  hands  4  days  ;  the  whole  crop  amounted  to 
372  bushels  ;  how  much  was  each  one's  share  ? 

5.  "William  Gallup  began  trade  January  1,  1856,  with  a 
capital  of  $3000,  and,  succeeding  in  business,  took  in  M.  H. 
Decker  as  a  partner  on  the  first  day  of  March  following,  with 


283  ANALYSIS. 

a  capital  of  $2000;  four  months  after  they  admittedJ.  New- 
man as  third  partner,  Avho  put  in  $1800  capital;  they  con- 
tinued their  partner.-^hip  until  April  1,  1858,  when  they  found 
tliat  $4388.80  had  been  gained  since  Jan.  1,  1856;  how 
much  was  each  one's  share  ?  (  §2106,  Gallup's. 

Ans.-\  $1300,  Decker's. 

(  $  982.80,  Newman's. 
6o  Two  persons  engaged  in  partnership  with  a  capital  of 
$5 GOO;  A's  capital  was  in  trade  8  months,  and  his  share  of 
the  profits  was  $560  ;  B's  capital  was  in  10  months,  and  his 
share  of  the  profits  was  $800 ;  what  amount  of  capital  had 
each  in  the  firm  ?  Ans.  A,  $2613.33^;  B,  $298G.6Gf. 

7.  A,  B,  and  C,  engaged  in  trade  with  $1930  capital;  A's 
money  was  in  3  months,  B's  5,  and  C's  7  ;  they  gained  $117, 
which  was  so  divided  that  a  of  A's  share  was  equal  to  ^-  of 
B's  and  to  ^  of  C's ;  how  much  did  each  put  in,  and  what 
did  each  gain?  (  A,  $700  capital ;  $26  gain. 

A)is.  ]B,$GS0       "       $39     " 
C,  $600       "       $52     « 

ANALYSIS. 

30®.  Analysis,  in  arithmetic,  is  the  process  of  solving 
problems  independently  of  set  rules,  by  tracing  the  relations 
of  the  given  numbers  and  the  reasons  of  the  separate  steps  of 
the  operation  according  to  the  special  conditions  of  each  question. 

397.  In  solving  questions  by  analysis,  we  generally  reason 
from  the  ffiven  number  to  tinity,  or  1,  and  then  from  unity,  or 
1,  ^o  the  required  number. 

398.  United  States  money  is  reckoned  in  dollars,  dimes, 
cents,andmills(180),one  dollar  being  uniformly  valued  in  all 
tlie  States  at  100  cents  ;  but  in  most  of  the  States  money  is 
sometimes  still  reckoned  in  pounds,  shillings,  and  pence. 

Note.  At  tho  time  of  the  adoption  of  our  decimal  currency  by 
Congress,  in  1786,  the  colonial  citrrenci/,  or  bills  of  credit,  issued  by  the 
colonies,  had  depreciated  in  value,  and  this  depreciation,  being  imequal 
in  the  different  colonies,  gave  rise  to  tho  different  values  of  the  State 
currencies ;  and  this  variation  continues  wherever  the  denominations  of 
bhillings  and  pence  are  in  use. 


ANALYSIS.  289 

S©9.    In  New  England,  Indiana,  \ 
Illinois,  Missouri,  Virginia,  Kentucky,  >  $1  m  6  s.  =  72  d. 
Tennessee,  Mississippi,  Texas, ^ 

New  York,  Ohio,  Micliigaii, $1  :=  8  s.  :=  06  d. 

New  Jersey,   Pennsylvania,  Dela- ) 

T.r       ,1  t$l  =  7s.  6d.=:00d 

Ware,  Maryland, ) 

South  Carolina,    Georgia, \  |l=4s.  8d.  =  56d. 

Canada,  Nova  Scotia, I  $1  =  5  s.  =  60  d. 


EXAMPLES    FOR    PRACTICE. 

1.    What  will  be  the  cost  of  42  bushels  of  oats,  at  3  shillings 
f-er  bushel,  New  England  currency  ? 

OPERATION. 

3 


42X3=  126  s.  0 

126 -^6  =$21     Or,    — 


121,  Ans. 


Analysis.  Since 
1  bushel  costs  3  shil- 
lings, 42  bushels  will 
cost  42  times  3  s.,  or 
42X3  =  126  s.;  and 
as  6  s.  make  1  dollar 

New  England  currency,  there  are  as  many  dollars  in  126  s.  as  6  is 

contained  times  in  126,  or  $21. 

2.   What  will   180  bushels  of  wheat  cost  at  9  s.  4d.  per 
bushel,  Pennsylvania  currency  ? 

.OPERATION. 

!  a'^02  Or, 


00    113 


$224 


28 
2 


$224,  Ans. 


Analysis.  Multi- 
plying the  number  of 
bushels  by  the  price, 
and  dividing  by  the 
value  of  1  dollar  re^ 
duced  to  paxce,  we 
we  have  $224.  Or, 
when  the  pence  in  the 
given  price  is  an  aliquot  part  of  a  shilling,  the  price  may  be  reduced 
to  an  improper  fraction  for  a  multiplier,  thus  :  ,9  s.  4  d.  rr  91  s.  =: 
2j?  s.,  the  multiplier.  The  value  of  the  dollar  being  7  s.  G  d.  z=l  1\  s. 
:=:  ^2*,  we  divide  by  ^,  as  in  the  operation. 

3.    What  will  be  the  cost  of  3  hhd.  of  molasses,  at  1  s.  3  d. 
per  quart,  Georgia  currency  ? 


R.p. 


18 


2r!:ii 


Pl> 


9, 


290  ANALYSIS. 

OPERATION. 

o  Analysis.     In  this  example  we  first 

003  reduce  3  lilid.  to  quarts,  by  multiplying 

y  by  G3  and  4,  and  then  multiply  by  the 

^  price,  either  reduced  to  pence  or  to  an 

^'^ improper   fraction,   and  divide    by  the 

405  00  value  of  1  dollar   reduced  to  the  same 
denomination  as  the  price. 

$£02.50 

4.  Sold  9  firkins  of  butter,  each  containing  56  lb.,  at  1  s.  G  d. 
per  pound,  and  received  in  payment  carpeting  at  6  s.  9  d.  per 
yard  ;  bow  many  yards  of  carpeting  would  pay  for  the  butter  ? 

OPERATION.  Analysis.     The  operation  in  this  is  similar 

0  to  the  preceding  examples,  except  that  we  di- 

5Q  vide  the  cost  of  the  butter  by  the  price  of  a 


$i 


,^2  ^mit  of  the  article  received  in  payment,  reduced 

^  to  the  same  denominational  unit  as  the  price 


112  yd.  °^  ^  "^"^^^  °^  ^^^  article  sold.    The  result  will  be 

the  same  in  whatever  currency. 

5.  "What  will    3    casks   of  rice    cost,   each  weidiinc    126 

7  DO 

pounds,  at  4  d.  per  pound,  South  Carolina  cuiTency  ?   Ans.  $27. 

6.  How  many  pounds  of  tea,  at  7  s.  per  pound,  must  be 
given  for  28  lb.  of  butter,  at  1  s.  7  d.  per  pound  ?     Ans.    6^. 

7.  Bought  2  casks  of  Catawba  Avine,  each  containing  72 
gallons,  for  $648,  and  sold  it  at  the  rate  of  10  s.  6  d.  per  quart, 
Ohio  currency  ;  how  much  was  my  whole  gain  ?     Ans.    $108. 

8.  What  will  be  the  expense  of  keeping  2  horses  3  weeks 
if  tlic  expense  of  keeping  1  horse  1  day  be  2  s.  6  d.,  Canada 
currency?  Ans.    $21. 

9.  How  many  days'  work,  at  6  s.  3  d.  per  day,  must  be 
given  for  20  bushels  of  apples  at  3  s.  per  bushel  ?     Ans.    9f . 

10.  Bought  160  11).  of  dried  fruit,  at  Is.  6  d.  a  pound,  in 
New  York,  and  sold  it  for  2  s.  a  pound  in  Philadelphia  ;  how 
much  was  my  wliole  gain  ?  Ans.   $12,663. 

11.  A  merchant  exchanged  434-  yards  of  clotli,  worth  10  s. 
G  d.  per  yard,  for  other  cloth  worth  8  s.  3  d.  per  yard ;  liow 
many  yards  did  he  receive  ?  Ans.   55  j^j. 


ANALYSIS.  291 

12.  What  will  be  the  cost  of  300  bushels  of  wheat  at  9  s. 
4  d.  per  bushel,  Michigan  cun-encj  ?  Aiis.    $350. 

13.  If  f  of  f  of  a  toa  of  coal  cost  $2|,  how  much  Avill  f  of 
6  tons  cost  ? 

OPERATION. 

$  j  1'2^  Analysis..     Since  |  of  f  =  ^|  of  a  ton  costs 

1^  I  o^4  $2|  =  $132,  1  ton  will  cost  28  times  ^^\  of  $^^2, 

w  .0  or  $^'-  X  ft  ;  and  -f  of  G  tons  z=i  ^  tons,  wiU 

'  ^  cost  -3^  times  f|  of  §-i/  — $16. 


t^lG,  Ans. 

14.  If  8  men  can  build  a  wall  20  ft.  Ions,  6  ft.  high,  and 
4  ft.  thick,  in  12  days,  working  10  hours  a  day,  in  how  many 
da/s  can  24  men  build  a  wall  200  ft.  long,  8  ft.  high,  and  6  ft. 
thick,  Avorking  8  hours  a  day  ? 

OPERATION. 

1^        $         10       J^00io       $       0 

-7  X  —  X  —  X X  — X— =  100  da. 

I       ^S        $  20         0       ^ 

Analysis.  Since  8  men  require  12  days  of  10  hours  each  to 
build  the  wall,  1  man  would  require  8  times  12  days  of  10  hours 
each,  and  10  times  (12  X  8)  days  of  1  hour  each.  To  build  a  wall 
1  ft.  long  would  require  ^  as  much  time  as  to  build  a  wall  20  ft. 
long  ;  to  build  a  wall  1  ft.  high  would  require  \  as  much  time  as  to 
build  a  wall  6  ft.  high;  to  build  a  Avail  1  ft.  thick,  \  as  much  time  as 
to  build  a  wall  4  ft.  tliick.  Now,  24  men  could  build  tliis  Mall  in  ^^ 
as  many  days,  by  Avorking  1  hour  a  day,  as  1  man  could  build  it, 
and  in  \  as  many  days  by  AA'orking  8  hours  a  day,  as  by  Avorking  1 
hour  a  day  ;  but  to  build  a  wall  200  ft.  long  Avould  require  200  times 
as  many  days  as  to  build  a  Avail  1  ft.  long ;  to  build  a  Avail  8  ft.  high 
Avould  require  8  times  as  many  days  as  to  build  a  Avail  1  ft.  high ; 
?nd  to  build  a  wall  6  ft.  thick  would  requh-e  6  times  as  many  days 
as  to  build  a  Avail  1  ft.  thick. 

15.  If  2  pounds  of  tea  are  AA'orth  11  pounds  of  coffee,  and 
3  pounds  of  coffee  are  worth  5  pounds  of  sugar,  and  18  pounds 
of  sugar  are  AA'orth  21  pounds  of  rice,  hoAv  many  pounds  of 
rice  can  be  purchased  Avith  12  pounds  of  tea? 


292  ANALYSIS. 

OPERATION.  Analysis.     Since  18  lb.  of  su- 

^1'7  gar  are  equal  in  value  to  21  lb.  of 

3^^     5  rice,  1  lb.  of  sugar  is  equal  to  ^^ 

jg      2]^  .         of  21  lb.  of  rice,  or  fi  =  |  lb.  of 

w      j/^s  rice,  and  o  lb.  of  sugar  'are  equal 

— — to  5  times  |-  lb.  of  rice,  or  ^^P-  lb. ; 

3  I  385  if  3  lb.  of  coffee  are  equal  to  5  lb. 

Ans.  T>8Mb.  °^  ^^°f"'  °^"  ^  ^^'  °^  "'^'^'  ^  ^^'  ^^ 

^      '  coffee  is   equal  to  -J-   of  ^q°-  lb.  of 

rice,  or  f|  lb.,  and  11  lb.  of  coffee  are  equal  to  11  times  ff  lb.  of 

rice,  or  -Y/  lb. ;  if  2  lb.  of  tea  are  equal  to  1 1  lb.  of  coffee,  or  -W^-  lb. 

of  rice,  1  lb  of  tea  is  equal  to  ^  of  -\\5.  lb.  of  rice,  or  -^g^i  lb.,  and  12 

lb.  of  tea  are  equal  to  12  times  -%\^  lb.  of  rice,  or  -SfA  lb.  =  128  A  lb. 

16.  If  16  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days?  Ans.  72. 

17.  If  $10i-  will  buy  4§  cords  of  wood,  how  many  cords 
can  be  bought  for  $24|  ?  Ans.  11. 

18.  Gave  52  bai-rels  of  potatoes,  each  containing  3  bushels, 
v.'orth  33^-  cents  a  bushel,  for  65  yards  of  cloth;  how  much 
was  the  cloth  worth  per  yard  ?  Ans.  $.80. 

19.  If  a  staff  3  ft.  long  cast  a  shadow  5  ft.  in  length,  Avhat 
is  the  height  of  an  object  that  casts  a  shadow  of  46f  ft.  at  the 
same  time  of  day  ?  Ans.  28  ft. 

20.  Three  men  hired  a  pasture  for  $63  ;  A  put  in  8  sheep 
74  months,  B  put  in  .12  sheep  4^  months,  and  C  put  in  15 
sheep  6|  months  ;  ho\t  much  must  each  pay  ? 

21.  If  7  bushels  of  wheat  are  worth  10  bushels  of  rye, 
and  5  bushels  of  rye  are  Avorth  14  bushels  of  oats,  and  6 
busliels  of  oats  are  worth  $3,  how  many  bushels  of  wheat  will 
S^30buy?  ^»s.  15. 

22.  If  $480  gain  $84  in  30  months,  what  capital  will  gain 
$21  in  15  months?  Ans.  $240. 

23.  How  many  yards  of  carpeting  §  of  a  yard  wide  are 
equal  to  28  yards  ^^  of  a  yard  wide?  Ans.  3 14-. 

24.  If  a  footman  travel  130  miles  in  3  days,  when  the  days 
are  14  hours  long,  in  how  many  days  of  7  hours  each  will  he 
travel  390  miles?  Ans.  IS. 


ANALYSIS.  293 

25.  If  6  men  can  cut  45  cords  of  wood  in  3  days,  liow 
many  cords  can  8  men  cut  in  9  days  ?  Ans.  180. 

26.  B's  age  is  IJ-  times  the  age  of  A,  and  C's  is  2y\j  times 
the  age  of  both,  and  the  sum  of  their  ages  is  93 ;  what  is  the 
age  of  each?  A>2s.     A's  age,  12  yrs. 

27.  If  A  can  do  as  much  work  in  3  days  as  B  can  do  in 
4i  days,  and  B  can  do  as  much  in  9  days  as  C  in  12  days, 
;>.ii<l  C  as  mucli  in  10  days  as  D  in  8,  how  many  days'  work 
done  by  D  are  equal  to  5  days'  done  by  A?  Ans.    8. 

28.  Tlie  hour  and  minute  hands  of  a  watch  are  together  at 
12  o'clock,  M. ;  when  will  they  be  exactly  together  the  third 
time  after  this  ? 

OPERATION.  Analysis.      Since 

12  X  TT  X  3  =  Oj\h..  the  minute  hand  pass- 

Ans.   3  h.  1 G  min.  21^9x  sec,  P.  M.  es  the  _  hour  hand  11 

times  in  12  hours,  if 
both  are  together  at  12,  the  minute  hand  will  pass  the  hour  hand 
tlie  first  time  in  Jy  of  12  hours,  or  1^^  hours  ;  it  wIU  pass  the  horn- 
hand  the  second  time  in  -^^  of  12  hours,  and  the  third  time  in  ^j  of 
12  hours,  or  3^^^  hours,  v.hich  would  occm:  at  16  min.  21^9.  ggc. 
past  3  o'clock,  P.  M. 

29.  A  flour  merchant  paid  $104  for  20  barrels  of  flour, 
giving  $9  for  first  quality,  and  $7  for  second  quality  ;  how 
many  barrels  were  there  of  each  ? 

OPERATION.  Analysis.     If  all  had  been 

S9  X  20  =  $180  ;  first  quality,  he  would  liave  paid 

cjgQ "^164  :z=  $16.  $180>  or  §16  more  tlian  he  did 

on is- 09  .  pay.      Every  barrel   of  second 

""         o"^  "  '1  quality  made  a  difference  of  $2 

16-1-2=     8  bbl.,  2d  quality.  j^^   ^j^g    ^^^^  .    j^g^^^   ^^^^.^   ^^.^^.g 

■^0 — 8  =  12  bbl.,  1st     «  as  many  barrels  of  second  qual- 

ity as  $2,  the  difference  in  the 
cost  of  one  barrel,  is  contained  times  in  $16,  &c. 

30.  A  boy  bought  a  certain  number  of  oranges  at  the  rate 
of  3  for  4  cents,  and  as  many  more  at  the  rate  of  5  for  8  cents  ; 
he  sold  them  again  at  the  rate  of  3  for  8  cents,  and  gained  on 
the  whole  108  cents;  how  many  oranges  did  he  buy  ? 


294.  ANALYSIS. 

oPERATiox.  Analysis.   For 

§   4"    f   ^^  if;   15  ^2=:  ff,  average  cost.         those    he    bought 
§  —  ff  =  If  =r  1 J  cts.,  gain  on  each.  at  the  rate  of  3  for 

108  -^li  =  90,  number  of  oranges.  ^  ^ents  he  paid  | 

of  a  cent  each,  and 
for  those  he  bought  at  the  rate  of  5  for  8  cents  he  paid  |  of  a  cent 
each;  and  -<-ff  — -4|  cents,  what  he  paid  for  1  of  each  kind, 
wliich  divided  by  2  gives  f  2  cents,  the  average  price  of^all  he  bought. 
He  sold  them  at  the  rate  of  3  for  8  cents,  or  |  cents  each ;  the  dif- 
ference between  the  average  cost  and  the  price  he  sold  them  for,  or 
f  —  if  =  If  —  H  cents,  is  the  gain  on  each ;  and  he  bought  as 
many  oranges  as  the  gain  on  one  orange  is  contained  times  in  the 
whole  gain,  &c. 

31.  A  man  bought  10  bushels  of  wheat  and  25  bushels  of 
corn  for  $30,  and  12  bushels  of  wheat  and  5  bushels  of  corn 
for  $20 ;  how  much  a  bushel  did  he  give  for  each  ? 

Analysis.  We  may  divide  or 
multiply  either  of  the  expressions 
by  such  a  number  as  will  render 
one  of  the  commodities  purchased, 
ahke  in  both  expressions.  In  this 
example  we  divide  the  first  by  5 
to  make  the  numbers  denoting 
the  corn  alike,  (the  same  result 
would  be  produced  by  multiply- 
ing the  second  by  5,)  and  we  have 
the  cost  of  2  bushels  of  wheat  and  o  bushels  of  corn,  equal  to  SO. 
Subtracting  this  from  12  bushels  of  wheat  and  5  bushels  of  corn,  which 
cost  820,  we  find  the  cost  of  10  bushels  of  wheat  to  be  $14  ;  there- 
fore the  cost  of  1  bushel  is  J^  of  $14,  or  $1.40.  From  any  one  of 
the  expressions  containing  both  wheat  and  corn,  we  readily  find  the 
cost  of  1  bushel  of  corn  to  be  64  cents. 

32.  A,  B,  and  C  agree  to  build  a  barn  for  $270.  A  and 
B  can  do  the  work  in  16  days,  B  and  C  in  13^  days,  and  A 
and  C  in  lly  days.  In  how  many  days  can  all  do  it  working 
together  ?  In  how  many  days  can  each  do  it  alone  ?  "What 
part  of  the  \y.\y  ought  each  to  receive  ? 


OPEKATION. 

W. 

C. 

1st  lot, 

10 

25 

$30 

2d    « 

12 

5 

$20 

1st    • 

• 

5  =  2 

5 

$6 

10. 

•     e     • 

.  $14 

1  bu.  W. 

= 

$1.40 

1  bu.  C. 

— 

$  .64 

_£- 

5 

4 

•   1    -' 

A-  = 

;20cla 

.,  C  alone. 

80 

^0  ■ 

SO 

so  — 

_9_ 

6_ 

_3_ 

.    1-|- 

•A-= 

261  d 

a.,  A   " 

80 

80 

80 

80 

6 

9 

7    . 

2 

;   1-^ 

2    

40  da. 

.,  B      " 

80 

80 

80 

8  0 

A 

X8|- 

=  *>■ 

the  part  of  the 

I  whole  C  did. 

3 

X8f 

.  3 

cc 

(( 

a 

A   « 

80 

9' 

0 

•80 

X8| 

1' 

u 

IC 

(1 

B   « 

ANALYSIS.  295 

OPERATION.  Analysis.    Since 
JL  :=  _5_.^  what  A  and  B  do  in  1  day.  A  and  B  can  do  tlie 
3_— -  6       "     BandC     "        "  Avoik in  16days,tlley 
^T_  ^r  -JL,     "    A  and  C     «        "  Can  do  ^q-  of  it  in  1 

■io  +  -s'V  + 10  =  80'  '"''^*  ^'  ^'  '^'"^  ^  ^°  ''^         '^''^y ' ,  ^    ^'^'■^  ^'   "^ 

2  dajs.  13|-  or  ^°-  days,  they 

.18  _^  2  z=  -T^g-,  what  A,  B,  and  C  do  in  1  day.  can  do  -^\  of  it  in  1 

1  _|_    9-  :rr  8|  days,  time  A,  B,  and  C,  will  do  the  day ;  A  and  C,  in  llf 

whole  work  together.  _    or  \°-  days,  they  can 

do  -/g  of  it  in  1  day. 
Then  A,  B,  and  C, 
by  working  2  days 
each,  can  do  -^q-\- 

_3 I 7_ 18  of  tiie 

40T^80  80  "*   *-  ^ 

work,  and  by  work- 
ing 1  day  each  they 
can  do  i  of  l|,  or  -^y 
$270  X  I  =  $120,  C"s  f-hare.  of  the  work  ;  and  it 

$270Xf  =  $90.    A's    "  will   take    them    as 

$270  X  7  =  $60,    E-s     '••  many  days  working 

together  to   do  the 
Avliole  work  as  -^^  is  contained  times  in  1,  or  8f  days. 

Now,  if  we  take  what  any  two  of  them  do  in  1  day  from  what  the 
three  do  in  1  day,  the  remainder  will  be  what  the  third  docs ;  we 
thus  find  that  A  does  -f^,  B  -^^,  and  C  -^^. 

Next,  if  we  denote  the  whole  work  by  1,  and  divide  it  by  the  part 
each  does  in  1  day,  we  have  thQ  number  of  days  that  it  will  take 
each  to  do  it  alone,  viz. :  A  26|  days,  B  40  days,  and  C  20  days. 
And  each  should  receive  such  a  part  of  $270  as  would  be  ex- 
pressed by  the  i^art  he  does  in  1  day,  multiplied  by  the  number  of 
days  he  works,  which  will  give  to  A  $90,  B  $60,  and  C  $120. 

33.  If  G  oranges  and  7  lemons  cost  33  cents,  and  12  oranges 
and  10  lemons  cost  54  cents,  what  is  the  price  of  1  of  each  ? 
A)is.    Oranges,  2  cents;  lemons,  3  cents. 

31.  If  an  army  of  1000  men  have  provisions  for  20  days, 
at  the  rate  of  18  oz,  a  day  to  each  man,  and  they  be  reinforced 
by  GOO  men,  upon  what  allowance  per  day  must  each  man  be 
put,  that  the  same  provisions  may  last  30  days  ?    Ans.    74-  oz. 

3'J.  There  are  54  bushels  of  grain  in  2  bins  ;  and  in  one  bin 
are  G  bushels  less  than  ^  as  much  as  there  is  in  the  other ; 
how  many  bushels  in  the  larger  bin  ?  Ans.    40. 


296  ANALYSIS. 

3G.  The  sum  of  two  numbers  is  20,  and  their  difference  is 
equal  to  ^  of  the  greater  number;  what  is  the  greater 
number?  Ans.    12. 

37.  If  A  can  do  as  much  work  in  2  days  as  C  in  3  days, 
and  B  as  mucli  in  5  days  as  C  in  4  days;  what  time  will  B 
require  to  execute  a  piece  of  w^ork  which  A  can  do  in  6 
weeks?  Ans.    11^  weeks. 

38.  How  many  5^ards  of  cloth,  f  of  a  yard  wide,  will  line 
oQ  yards  1;^  yards  wide  ?  Ans.    GO. 

39.  How  many  sacks  of  coffee,  each  containing  104  lbs, 
at  10  d.  per  pound  N.  Y.  currency,  will  pay  for  80  yards  of 
broadcloth  at  $3|-  per  yard  ?  Ans.    24. 

40.  A  person,  being  asked  the  time  of  daj',  replied,  the  time 
past  noon  is  equal  to  -i  of  the  time  to  midnight ;  v/hat  was 
the  hour  ?  Ans.    2,  P.  M. 

41.  A  market  woman  bought  a  number  of  peaches  at  the 
rate  of  2  for  1  cent,  and  as  many  more  at  the  rate  of  3  for  1 
cent,  and  sold  them  at  the  rate  of  5  for  3  cents,  gaining  55 
cents  ;  how  many  peaches  did  she  buy  ?  Ans.   300. 

42.  A  can  build  a  boat  in  18  days,  working  10  hours  a  day, 
and  B  can  build  it  in  9  days,  working  8  hours  a  day ;  in  liow 
many  days  can  both  together  build  it,  working  6  hours  a  day  ? 

43.  A  man,  after  spending  ^  of  his  money,  and  ^  of  the 
remainder,  had  $10  left ;    how  much  had  he  at  first  ? 

44.  If  30  men  can  perform  a  piece  of  work  in  1 1  days,  how 
many  men  can  accomplish  another  piece  of  work,  4  times  as 
large,  in  J-  of  the  time  ?  Ans.    600. 

45.  If  1G|-  lb.  of  coffee  cost  $3|,  how  much  can  be  bought 
for  $1.25?  Ans.    Glib. 

46.  A  man  engaged  to  write  for  20  days,  receiving  $2.50 
for  every  day  he  labored,  and  forfeiting  $1  for  every  day  he 
was  idle ;  at  the  end  of  the  time  he  received  $43  ;  liow  many 
days  did  he  labor  ?  Ans.    18. 

47.  A,  B,  and  C  can  perform  apiece  of  work  in  12  hours; 
A  and  B  can  do  it  in  16  hours,  and  A  and  C  in  18  hours; 
what  part  of  the  work  can  B  and  C  do  in  9|  hours  ?     Ans.  f. 


ALLIGATION   MEDIAL,  e  .-  297 

ALLIGATION". 

^€?>0.  Alligation  treats  of  mixing  or  compounding  two  or 
more  ingredients  of  different  values.  It  is  of  two  kinds  —  Alli- 
gation Medial  and  Alligation  Alternate. 

401 .  Alligation  Medial  is  tlie  process  of  finding  the  aver- 
age price  or  quality  of  a  compound  of  several  simple  ingredi- 
ents whose  prices  or  qualities  are  known. 

1.  A  miller  mixes  40  bushels  of  rye  worth  80  cents  a 
bushel,  and  25  bushels  of  corn  Avorth  70  cents  a  bushel,  with 
15  bushels  of  wheat  worth  §1.50  a  bushel;  what  is  the  value 
of  a  bushel  of  the  mixture  ? 

OPF.RATIOX.  AxALYSIS^  Since   40  bushels 

80  X  40  =  $32.00  of  rye  at  80  cents  a  bushel  is 

70  X  25  rr:     17.50  worth  $32,  and  2.J  bushels  of  corn 

1  50  y  15  =     22  50  ^^    "'^  cents    a  bushel   is   worth 

—       $17.50,  and  15  bushels  of  wheat 

80       )  72.00  at  $1.50  a  bushel  is  worth  $22.50, 

fit;  nn     A„g  therefore  the  entire  mixture,  con- 
sisting of   80   bushels,  is  worth 

$72,   and   one  bushel   is  worth  J^   of  $72,  or   72  -I-  SO  :=  $.90. 
Hence  the  following 

Rule.  Divide  the  entire  cost  or  value  of  the  ingredients 
bg  the  sum  of  the  simples. 

EXAMPLES    FOR    rRACTICE. 

2.  A  wine  merchant  mixes  12  gallons  of  wine,  at  %l  per 
gallon,  with  5"  gallons  of  brandy  worth  $1.50  per  gallon,  and 
3  gallons  of  water  of  no  value ;  what  is  the  worth  of  one  gal- 
lon of  tlie  mixture  ?  Ans.    $.975. 

3.  An  innkeeper  mixed  13  gallons  of  water  with  52  gallons 
of  brandy,  which  cost  him  §1.25  per  gallon  ;  what  is  the  value 
of  1  gallon  of  the  mixture,  and  what  his  profit  on  the  sale  of 
the  whole  at  C|-  cents  per  gill  ?    Ans.    %l  a  gallon  ;  $65  profit. 

4.  A  grocer  mixed   10   pounds  of  sugar  at  8  cts.  with  12 

pounds  at  9  cts.  and  IG  pounds  at  11  cts.,  and  sold  the  mixture 

at  10  cents  per  pound;  did  he  gain  or  lose  by  the  sale,  and 

hov/much?  A71S.    He  gained  16  cts. 

13* 


293  ALLIGATION  ALTERNATE. 

5.  A  gi'ocer  bought  7i  dozen  of  eggs  at  12  cents  a  dozen, 
8  dozen  at  10^  cents  a  dozen,  9  dozen  at  11  cents  a  dozen, 
and  101  dozen  at  10  cents  a  dozen.  He  sells  them  so  as  to 
make  50  per  cent,  on  the  cost  j  how  much  did  he  receive  per 
dozen?  Alls.    10}   cents. 

6.  Bought  4  cheeses,  each  weighhig  50  pounds,  at  13  cents 
a  pound;  10,  weighing  40  jiounds  each,  at  10  cents  a  pound; 
and  24,  weighing  25  pounds  each,  at  7  cents  a  pound;  I  sold 
the  whole  at  an  average  price  of  9i  cents  a  pound;  how  much 
Vt'as  my  whole  gain  ?  Ans.    $G. 

4:®?5.  Alligation  Alternate  is  the  process  of  finding  the 
proportional  quantities  to  be  taken  of  several  ingredients, 
v.'hose  prices  or  qualities  are  known,  to  form  a  mixture  of  a 
required  price  or  quality. 

CASE    I. 

4® 3.  To  find  the  proi^ortional  quantity  to  be  used 
of  each  ingredient,  when  the  mean  price  or  quality  of 
the  mixture  is  given. 

1.  What  relative  quantities  of  timothy  seed  worth  $2  a 
busliel,  and  clover  seed  worth  $7  a  bushel,  must  be  used  to 
form  a  mixture  worth  $5  a  bushel  ? 

OPERATION.  Analysis.     Since  on  every  in- 


1 


2  )  gradient  used  whose  price  or  qnal- 

3  j  ^"^*  ity  is  less  than  the  mean  rate  there 
will  be  a  gaii>,  and  on  every  in- 
gredient whoso  price  or  quality  is  greater  than  the  mean  rate 
there  will  be  a  loss,  and  since  the  gains  and  losses  must  be  exactly 
equal,  the  relative  quantities  used  of  each  shoidd  be  such  as  repre- 
sent the  unit  of  value.  By  selling  one  bushel  of  timothy  seed  worth 
$2,  for  $0,  there  is  a  gain  of  $3  ;  and  to  gain  $1  would  require  J  of 
a  buslicl,  which  we  place  opposite  the  2.  By  selling  one  bushel  of 
clover  seed  worth  $7,  for  $.5,  there  is  a  loss  of  $2  ;  and  to  lose  $1 
would  require  ^  of  a  bushel,  wliicli  we  place  opposite  tlie  7. 

In  every  case,  to  find  the  unit  of  value  we  must  divide  $1  by  tlie 
gain  or  loss  per  bushel  or  pound,  &-c.  Hence,  if,  every  time  we  take 
!^  of  a  bushel  of  timothy  seed,  we  take  i  of  a  bushel  of  clover  seed, 
the  gain  and  loss  will  be  exactly  equal,  and  we  shall  have  \  and  -^ 
for  the  proportional  quantities. 


i 


ALLIGATION   ALTERNATE. 


299 


OPERATION. 


1 

2 

o 

4 

0 

3 

1 

4 

4 

4 

X 

'J 

1 

1 

7 

1 

2 

2 

I  10 

1 

o 

3 

If  we  wish  to  express  the  proportional  numbers  in  integers,  we 
may  reduce  these  fractions  to  a  common  denominator,  and  use  their 
numerators,  since  fractions  having  a  common  denominator  are  to 
each  other  as  their  numerators.  ( 365)  thus,  \  and  i  are  equal  to 
I  and  |,  and  the  proportional  quantities  are  2  bushels  of  timothy 
seed  to  3  bushels  of  clover  seed. 

2.  Wliat  proportions  of  teas  worth  respectively  3,  4,  7  and 
10  shillings  a  pound,  must  be  taken  to  form  a  mixture  worth 
6  shillings  a  pound  ? 

Analysis.  To  preserve  the 
equalit}'  of  gains  and  losses,  we 
must  always  compare  two  prices 
or  simples,  one  greater  and  one 
less  than  the  mean  rate,  and 
treat  each  pair  or  couplet  as  a 
separate  example.  In  the  given 
example  we  form  two  couplets, 

and  may  compare  either  3  and  10,  4  and  7,  or  3  and  7,  4  and  10. 
A7e  find  that  ^  of  a  lb.  at  3  s.  must  be  taken  to  gain  1  shilling, 

and  ^  of  a  lb.  at  10  s.  to  lose  1  shilling  ;  also  -|  of  a  lb.  at  4  s.  to  gain 

1  shilling,  and  1  lb.  at  7  s.  to  lose  1  shilling.  These  proportional 
numbers,  obtained  by  comparing  the  two  couplets,  are  placed  in 
columns  1  and  2.     If,  now,  we  reduce  the  num])ers  in  columns  1  and 

2  to  a  common  denominator,  and  use  their  numerators,  we  obtain 
the  integral  numbers  in  columns  3  and  4,  which,  being  arranged  in 
column  o,  give  the  proportional  quantities  to  be  taken  of  each.* 

It  will  be  seen  that  in  comparing  the  simples  of  any  couplet,  one 
of  which  is  greater,  and  the  other  less  than  the  mean  rate,  the  pro- 
portional number  finally  obtained  for  either  term  is  the  difference 
between  the  mean  rate  and  the  other  term.  Thus,  in  comparing  3 
and  10,  the  proportional  number  of  the  former  is  4,  which  is  the 
difference  between  10  and  the  mean  rate  6  ;  and  the  proportional 
number  of  the  latter  is  3,  which  is  the  difference  between  3  and  the 
mean  rate.  Tlie  same  is  true  of  every  other  couplet.  Hence,  when 
the  simples  and  the  mean  rate  are  integers,  the  intermediate  steps 
taken  to  obtain  the  final  proportional  numljers  as  in  columns  1,  2,  3, 
and  4,  may  be  omitted,  and  the  same  results  readily  found  by  taking 
the  difference  between  each  simple  and  the  mean  rate,  and  placing 
it  opposite  the  one  with  which  it  is  compared. 


*  Prof.  A.  B.  Oanfipld,  of  Oneida  Conference  Seminary,  N.  Y.,  used  this  method  of 
Allisation,  essentialij'.  in  the  instruction  of  his  classes  as  early  as  1846,  and  he  was 
duubtless  the  author  of  it. 


oOO  ALLIGATION   ALTERNATE. 

From  the  foregoing  esamples  and  analyses  we  derive  the  following 

Rule.  I.  Write  the  several  prices  or  qualities  in  a  column, 
and  the  mean  price  or  quality  of  the  mixture  at  the  left. 

II.  Form  couplets  by  comparing  any  price  or  quality  less, 
vjith  one  that  is  greater  than  the  mean  rate,  placing  the  part 
ichich  inust  he  used  to  gain  1  of  the  mean  rate  opposite  the  less 
simple,  and  the  part  that  must  be  used  to  lose  1  opposite  the 
grecder  simple,  and  do  the  same  for  each  simple  in  every  couplet. 

III.  If  the  proportional  numbers  are  fractioncd,  they  may  be 
reduced  to  integers,  and  if  tioo  or  more  stand  in  the  same  hori- 
zontal line,  they  must  he  added ;  the  final  results  will  he  the  pro- 
portional quantities  required. 

Notes.  1.  If  the  numbers  in  any  couplet  or  column  have  a  com- 
mon factor,  it  may  bs  rejected. 

2.  We  may  also  multiply  the  numbers  in  any  couplet  or  coliunn  by 
any  multiplier  we  choose,  without  affecting  the  equality  of  the  gains 
and  losses,  and  thus  obtain  an  indefinite  number  of  results,  any  one  of 
which  being  taken  will  give  a  correct  ihial  result. 

EXAMPLES    FOR   PRACTICE. 

3.  A  grocer  has  sugars  worth  10  cents,  11  cents,  and  14 
cents  per  pound;  in  what  proportions  nmy  he  mix  them  to 
form  a  mixture  worth  12  cents  per  pbund? 

Ans.  1  lb.  at  10  cts.,  and  2  lbs.  at  11  and  14cts. 

4.  What  proportions  of  water  at  no  value,  and  wine  worth 
$1.20  a  gallon,  muit  be  used  to  form  a  mixture  worth  90  cents 
a  gallon  ?  Ans.    1  gal.  of  water  to  3  gals,  of  wine. 

5.  A  farmer  had  sheep  worth  $2,  $2^,  $3,  and  $4  per 
head;  what  number  could  he  sell  of  each,  and  realize  an 
average  price  of  §2^  per  head  ? 

A^ns.  3  of  the  1st  kind,  and  1  each  of  the  2d  and  3d, 
and  5  of  the  4th  kind, 

6.  What  relative  quantities  of  alcohol  80,  84,  87,  94,  and 
96  per  cent,  strong  must  be  used  to  form  a  mixture  90  per 
cent,  strong  ? 

Ans.  G  of  the  first  two  khids,  fouj'  of  the  3d,  3  of  the  4th, 
and  16  of  the  5tli, 


OPERATION. 

30 

3^ 

4 

4 

45 

t\ 

8 

8 

8i 

A 

^\ 

5 

5 

10 

ALLIGATION   ALTERNATE.  301 

CASE  II. 

^®4.  When  tlic  quantity  of  one  of  the  simples  is 
limited. 

1.  A  miller  has  oats  worth  30  cents,  corn  worth  45  cents, 
raid  barley  worth  84  cents  per  bushel ;  he  desires  to  form  a 
mixture  worth  60  cents  per  bushel,  and  which  shall  contain  40 
bushels  of  corn  ;  how  many  bushels  of  oats  and  barley  must 
he  take  ? 

Analysis.  By 
20   )  the  same  process 

60  ^  45     "      yL  8       8    40   V  Am.      ^^  "^  Case  I  we 

rQ    \  -find   the    propor- 

tional quantities 
cf  each  to  be  4  bushels  of  oats,  8  of  corn,  and  10  of  barley.  But 
we  wish  to  use  40  bushels  of  corn,  which  is  5  times  the  propor- 
tional number  8,  and  to  preserve  the  equality  of  gain  and  loss  M'e 
must  take  5  times  the  proportional  quantity  of  each  of  the  other 
simples,  or  5  X  4  =r  20  bushels  of  oats,  and  5  X  10  =r  50  bushels 
of  barley.     Hence  the  following 

Rule.  Find  the  proportional  quantities  as  in  Case  I. 
Divide  the  given  quantity  by  the  proportional  quantity  of  the 
same  ingredient,  and  multipty  each  of  the  other  proportional 
quantities  hy  the  quotient  thus  obtained. 

EXAMPLES    FOR    PRACTICE. 

2.  A  merchant  has  teas  worth  40,  60,  75,  and  90  cents  per 
pound  ;  how  many  pounds  of  each  must  he  use  with  20  pounds 
of  that  Avorth  75  cents,  to  form  a  mixture  at  80  cents. 

Ans.  20  lbs.  each  of  the  first  three  kinds,  and  130  lbs.  of 
the  fourth. 

3.  A  farmer  bought  24  sheep  at  $2  a  head  ;  how  many 
must  he  buy  at  $3  and  $5  a  head,  that  he  may  sell  the  whole 
at  an  average  price  of  $4  a  head,  without  loss  ? 

Ans.    24  at  $3,  and  72  at  $5. 

4.  IIov/  much  alcohol  worth  60  cents  a  gallon,  and  how 
much  v.'ater,  must  be  mixed  with  180  gallons  of  rum  worth 
SI. 30  a  gallon,  that  the  mixture  may  be  worth  90  cents  a 
gallon  ?  Ans.    60  gallons  each  of  alcohol  and  water. 


^Q2 


ALLIGATION   ALTERNATE. 


5.  How  many  acres  of  land  Avorth  35  dollars  an  acre  must 
be  added  to  a  farm  of  75  acres,  -vvortli  $50  an  acre,  that  the 
average  value  may  be  $iO  an  acre  ?  Ajis.    150  acres. 

6.  A  merchant  mixed  80  pounds  of  sugar  -worth  G^  cents 
per  pound  with  some  worth  8^  cents  and  10  cents  per  pound, 
so  that  the  mixture  was  worth  71  cents  per  pound  ;  how  much 
of  each  kind  did  he  use  ? 

CASE    III. 

4:©^,  When  the  quantity  of  the  whole  compound  is 
limited. 

1.  A  grocer  has  sugars  wortli  G  cents,  7  cents,  12  cents, 
and  13  cents  per  pound.  He  wishes  to  make  a  mixture  of 
120  pounds  worth  10  cents  a  pound;  how  many  pounds  of 
each  kind  must  he  use  ? 

OPERATION.  Analysis.    By  Case 

I  we  find  the  propor- 
tional quantities  of  each 
to  be  3  lbs.  at  6  cts.,  2 
lbs.  at  7  cts.,  3  lbs  at  12 
cts.,  and  4  lbs.  at  13  cts. 
By  adding  the  propor- 
tional quantities,  we  find 
that  the  mixture  would  be  but  12  lbs.  while  the  required  mixture  is 
120,  or  10  times  12.  If  the  whole  mixture  is  to  be  10  times  as  much 
as  the  sum  of  the  proportional  quantities,  then  the  quantity  of  each 
simple  used  must  be  10  times  as  much  as  its  respective  proportion- 
al, wliich  would  require  30  lbs.  at  6  cts.,  20  lbs.  at  7  cts.,  30  lbs.  at 
12  cts.,  and  -JO  llis.  at  13  cts.    Hence  Ave  deduce  the  foUowhig 

Rule.  Find  the  proportional  numbers  as  in  Case  I.  Di- 
vide the  given  quantity  hj  the  sum  of  the  proportional  qvan- 
tities,  and  multipuly  each  of  the  proportional  quantities  hy  the 
quotient  thus  obtained. 

EXAMPLES   FOR   PRACTICE. 

2.  A  farmer  sold  170  sheep  at  an  average  price  of  14 
shillings  a  head  ;  for  some  he  received  9  s.,  for  some  12  s.,  for 
some  18  s.,  and  for  others  20?.;  how  many  of  each  did  lie 
sell  ?     Ans.    GO  at  9  s.,  40  at  12  s.,  -20  at  18  s.,  and  50  at  20  s. 


\h 

1 

1 

3 

1 

1 

3 

30 

h 

2 

2 

20 

i 

3 

3 

30 

h 

4 

m 

4 

40 

12 

120 

INVOLUTION.  '  3u3 

3.  A  jeweler  melted  together  gold  16,  18,  21,  and  24 
cariits  line,  so  as  to  make  a  compound  of  51  ounces  22  carats 
fine  ;  how  much  of  each  sort  did  he  take  ?  Ans.  6  ounces 
each  of  the  first  three,  and  33  ounces  of  the  last. 

4.  A  man  bought  210  bushels  of  outs,  corn,  and  wheat,  and 
paid  for  the  whole  $178.50  ;  for  the  oats  he  paid  $^,  for  the 
corn  $3,  and  for  the  wheat  $1^  per  bushel ;  how  many  bush- 
els of  each  kind  did  he  buy  ?  Ans.  78  bushels  each  of  oats 
and  corn,  ainl  54  bushels  of  wheat. 

5.  A,  B,  and  C  are  under  a  joint  contract  to  furnish  6000 
bushels  of  corn,  at  48  cts.  a  bushel  ;  A's  corn  is  worth  45  cts., 
B's  51  cts.,  and  C's  54  cts.  ;  liovv  many  bushels  must  each  put 
into  tlie  mixture  that  the  contract  may  be  fulfilled  ? 

6.  One  man  and  3  boys  received  $84  for  56  days'  labor;  the 
man  received  $3  per  day,  and  the  boys  SJ-,  $f,  and  $lf  re- 
spectively ;  how  many  days  did  each  labor  ?  Ans.  Tlie  man 
16  days,  and  the  boys  24,  4,  and  12  days  respectively. 

INVOLUTION. 

406,  A  Power  is  the  product  arising  from  multiplying  a 
number  by  itself,  or  repeating  it  several  times  as  a  factor ; 
thus,  in  2  X  2  X  2  =  8,  the  product,  8,  is  a  power  of  2. 

4®T.  The  Expongnt  of  a  power  is  the  number  denoting 
how  many  times  the  factor  is  repeated  to  produce  the  poAver, 
and  is  written  above  and  a  little  tc  the  right  of  the  factor;  thus, 
2  X  2  X  2  is  written  2-^,  in  which  3  is  the  exponent.  Exponents 
likewise  give  names  to  the  powers,  as  will  be  seen  in  the 
followimr  illustrations : 

3  =  3^  =     3,  the  first  power  of  3  ; 

3X3  =  3-  =     0,  the  second  power  of  3 

3X3X3  —  33  =:  27,  the  third  power  of  3. 

4:'?)8<.    The  Square  of  a  number  is  its  second  power. 
4:®S&.    The  Cube  of  a  number  is  its  third  power. 
410,    Involution  is  the  process  of  raising  a  number  to  a 
given  power. 


304  EVOLUTION. 

411.  A  Perfect  Power  is  a  number  that  can  be  exactly 
produced  by  the  involution  of  some  number  as  a  root ;  thus,  25 
and  32  are  perfect  powers,  since  25  =  5  X  5,  and  32  =  2  X 
2X2X2X2. 

1.    What  is  the  cube  of  15? 

OPERATION.  Analysis.    We  multiply 

15  X  15  X  15  =:  3375.    Ans.         ^^  ^J'  l'^'  ^"^^  the  product 

by  15,  and  obtain  o375, 
which  is  the  3d  power,  or  cube  of  15,  since  15  has-been  taken  3 
times  as  a  factor.     Hence,  we  have  the  following 

Rule.  Multipli/  the  number  hj  itself  as  many  limes,  less  1, 
as  there  are  units  in  the  exponent  oj  the  required  power. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  square  of  25  ?  Ans.    625. 

3.  What  is  the  square  of  135  ?  Ans.    18225. 

4.  Wliat  is  the  cube  of  72  ?  Ans.   373248. 

5.  Yfhat  is  the  4th  power  of  24?  Ans.   331776. 
G.  Eaise  7.2  to  the  third  power.  Ans.    373,248. 

7.  Involve  LOG  to  the  4th  power.       Ans.    1.2G247696. 

8.  Involve   12  to  the  5th  power.       Ans.    .0000248832. 

9.  Involve  1.0002  to  the  2d  power.     Ans.  1.00040004. 

10.  What  is  the  cube  of  |  ? 

OPERATION. 

2        2        2       2X2X2        2^        8 

5        5        5~5X5X5~  53~125 

It  is  evident  from  the  above  operation,  that 
A  common  fraction  may  he  raised  to  any  poiocr,  hy  raising 
each  of  its  terms,  separately,  to  the  required  power. 

11.  What  is  the  square  of  f  ?  Ans.   -^^. 

12.  Wliat  is  the  cube  of  -ij  ?  Ans.   f  4-f  f 

13.  Raise  2 4f  to  the  2d  power.  Ans.    61 2^-^ 

EVOLUTION. 

4B2,  A  Soot  is  a  factor  repeated  to  produce  a  power; 
thus,  in  the  expression  5X5X5=:  125,  5  is  the  root  from 
whicli  the  power,  125,  is  produced. 


SQUARE   ROOT,  ?05 

4:13.  Evolution  is  the  process  of  extracting  the  root  of  a 
number  considered  as  a  power,  and  is  the  reverse  of  Involution. 

414.  The  Eadical  Sign  is  tlie  character,  V)  which,  placed 
before  a  number,  denotes  that  its  root  is  to  be  extracted. 

415.  The  Index  of  the  root  is  the  figure  placed  above  the 
radical  sign,  to  denote  what  root  is  to  be  taken.  When  no 
index  is  written,  the  index  2  is  always  understood. 

416.  A  Surd  is  the  indicated  root  of  an  imperfect  power. 
4tT.    Roots  are  named  from  the  coi-responding  powers,  as 

will  be  seen  in  the  following  illustrations : 

The  square  root  of  9  is  3,  wa-itten  ■\/d  z=  3. 
The  cube  root  of  27  is  3,  written  ^27  =  3. 
The  fourth  root  of  81  is  3,  written  ^^81  =  3. 

418.  Any  number  whatever  may  be  considered  a  power 
•whose  root  is  to  be  extracted  ;  but  only  the  perfect  powers  can 
have  exact  roots. 

SQTJAHE  EOOT. 

419.  The  Square  Eoot  of  a  number  is  one  of  the  two 
equal  factors  that  produce  the  number  ;  thus  the  square  root 
of  49  is  7,  for  7  X  7  =  49. 

422®.  In  extracting  the  square  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  given  number 
and  its  square  root.  The  law  governing  this  relation  is  exhib- 
ited in  the  following  examples  :  — 


Koots. 

Squares. 

Roots. 

Squares. 

1 

1 

1 

1 

9 

81 

10 

1,00 

99 

98,01 

100 

1,00,00 

999 

99,80,01 

1000 

1,00,00,00 

From  these  examples  we  perceive 

1st.  That  a  root  consisting  of  1  place  may  have  1  or  2 
places  in  the  square. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root 
adds  2  places  to  the  square.     Hence, 


306  EVOLUTION. 

If  we  point  off  a  number  into  two-fg^ire  periods,  commen- 
cing at  the  rigid  hand,  the  number  of  full  periods  and  the  left 
hand  full  or  partial  period  will  indicate  the  number  of  places 
in  the  square  root  ;  the  highest  period  answering  to  the  highest 
figure  of  the  root. 

4:*2i,  1.  What  is  the  length  of  one  side  of  (i  square  plat 
containing  an  area  of  5417  sq.  ft.  ? 

orERATiox.  Analysis.     Since  the  given  figin-e  is 

54,17  I  73.6  a  square,  its  side  will  be  the  square  root 

49  of  its  area,  which  we  will  proceed  to  com- 

pute.    Pointin":  off  the  given  number,  the 

i.-±\j      tji.t  2  periods  show  that  there  will  be  two  in- 

1 4o      42  J  tcgral  figures,  tens  and  units,  in  the  root. 

14G.0     88.00  ^^^^  ^^"^  ^^  *^^  ^'°'^^  must  be  exti-acted 

1  4  r  G      87  or  from  the  first  or  left  hand  period,  54  hun- 

dreds.     The  greatest  square  in  54  hun- 

4  dreds  is  49  hundreds,  the  square  of  7  tens  ; 

we  therefore  write  7  tens  in  the  root,  at 
the  right  of  the  given  number. 
Since  the  entire  root  is  to  be  the  side  of  a  square,  let  us  form  a 
Fii;  I.  square  (Fig.  I),  the  side  of  which  is  70  feet  long. 

The  area  of  this  square  is  70  X  70  ziz  4900'sq.  ft., 
which  we  subtract  from  the  given  number.  This 
is  done  in  the  operation  by  subtracting  the 
square  number,  49,  from  the  first  period,  54, 
and  to  the  remainder  bringing  down  the  sec- 
ond period,  making  the  entire  remainder  517. 
If  we  now  enlarge  our  square  (Fig.  I)  by  the  addition  of  517 
squai'e  feet,  in  such  a  manner  as  to  preserve  the  square  form,  its 
size  will  be  that  of  the  required  square.  To  preserve  the  square 
form,  the  addition  must  be  so  made  as  to  extend  the  square  equally 
in  two  directions  ;  it  will  therefore  be  composed  of  2  oblong  figures 
at  the  sides,  and  a  little  square  at  the  corner  (Fig.  II).  Now,  the 
width  of  this  addition  will  be  the  additional  length  to  the  side  of  the 
square,  and  consequently  the  next  figure  in  the  root.  To  find  width 
we  divide  square  contents,  or  area,  by  length.  But  the  length  of 
one  side  of  the  liy;le  square  cannot  be  found  till  the  width  of  the 
addition  be  determined,  because  it  is  equal  to  this  widtli.  We  mIU 
tlierefore  add  the  lengths  of  the  2  oblong  figures,  and  the  sum  will 
be  sufficiently  near  the  whole  length  to  be  used  as  a  trial  divisor. 


SQUARE   ROOT. 


307 


Fig.  II. 


7-. 

3 

70 

o 

70 


Trial  Divisor  =  140 


Complete  Divisor  =  143 


Each  of  the  oblong  figures  is  equal  in  length  to  the  side  of  the 

square  first  formed  ;  and  their  united  length 
is  70  +  70  =  140  ft.  (Fig.  III).  This  num- 
ber is  obtained  in  the  operation  by  doubling 
the  7  and  annexing  1  cipher,  the  result  being 
written  at  the  left  of  the  dividend.  Dividing 
517,  the  area,  by  140,  the  approximate  length, 
we  obtain  3,  the  probable  width  of  the  addi- 
tion, and  second  figure  of  the  root.  Since  3  is 
also  the  side  of  the  little  square,  wo  can  now 
find  the  entire  length  of  the  addition,  or  the  complete  divisor,  which 

is  70 -f  70 -|- 3  =  143  (Fig.  III). 
'^'      ■  This  number  is  found  in  the  oper- 

ation by  adding  3  to  the  trial  di- 
visor, and  writing  the  result  un- 
derneath. Multiplying  the  com- 
plete divisor,  143,  by  the  trial 
quotient  figure,  3,  and  subtracting 
tlie  product  from  the  dividend,  Ave 
obtain  another  remainder  of  88  square  feet.  With  this  remainder, 
for  the  same  reason  as  before,  we  must  proceed  to  make  a  new 
enlargement ;  and  mx-  bring  down  two  decimal  cii)hers,  because  the 
next  figure  of  the  root,  being  tenths,  its  square  will  l^e  hundredths. 
The  trial  divisor  to  obtain  the  width  of  this  new  enlargement, 
or  the  next  figure  in  the  root,  will  be,  for  the  same  reasgn  as 
before,  twice  73,  the  root  already  found,  with  one  cipher  annexed. 
But  since  the  7  has  already  been  doubled  in  the  operation,  we  have 
only  to  double  the  last  figure  of  the  complete  divisor,  143,  and 
annex  a  cipher,  to  obtain  the  new  trial  divisor,  146.0.  Dividing,  we 
obtain  .6  for  the  trial  figure  of  the  root ;  then  proceeding  as  before, 
we  obtain  146.6  for  a  complete  divisor,  87.96  for  a  product ;  and 
there  is  still  a  remainder  of  .04.  Hence,  the  side  of  the  given 
square  plat  is  73.6  feet,  nearly.  From  this  example  and  analysis 
we  deduce  the  following 

Rule.  I.  Point  off  the  given  numher  into  periods  of  two 
ff/ures  each,  counting  from  laiifs  place  toward  the  left  ajid 
riglt  t. 

II.  Find  the  greatest  square  numher  in  thelefl  hand  period, 
and  ivrite  its  root  for  the^,  first  figure  in  the  root ;  subtract  the 
square  numher  from  the  left  hand  period,  and  to  the  remainder 
^ring  down  the  next  period  for  a  dividend. 


i508  EVOLUTION". 

III.  At  (lie  Icji  of  the  dividend  lori'e  twice  the  jirst  figure  of 
the  root,  and  annex  one  ciphe7',fo7'  a  trial  divisor  ;  divide  the 
dividend  hy  the  trial  divisor,  and  icrite  the  quotient  for  a  trial 
figure  in  the  root. 

IV.  Add  the  trial  figure  of  the  root  to  the  trial  divisor  for  a 
complete  divisor ;  midtiply  the  complete  divisor  Ig  the  trial 
figure  in  the  root,  and  subtract  the  product  from  the  dividend, 
and  to  the  remainder  hring  doivn  the  next  period  for  a  new 
dividend. 

V.  To  the  last  complete  divisor  add  the  last  figure  of  the 
foot,  and  to  the  sum  annex  one  cipher,  for  a  new  trial  divisor, 
tvilh  which  proceed  as  before. 

Notes.  1.  If  at  anytime  the  product  be  greater  than  the  dividend, 
diminish  the  trial  figure  of  the  root,  and  correct  the  erroneous  -work. 

2.  If  a  cipher  occur  in  the  root,  annex  another  cipher  to  the  trial 
di-visor,  and  another  period  to  the  dividend,  and  proceed  as  before. 

EXAMPLES   FOR   PRACTICE. 

2.    What  is  the  square  root  of  406457.2516? 

OPERATIOX. 

40,64,57.25,16  |  637.54,  Ans. 


36 

Trial    divisor, 

120 

464 

Completo  " 

123 

369 

Trial         " 

1260 

9557 

Comiilete  " 

1267 

8869 

Trial         " 

1274.0 

688.25 

Complete  " 

1274.0 

637.25 

Trial           « 

1275.00 

51.0016 

Completo  " 

1275.04 

51.0016 

Notes.  3.  The  decimal  points  in  the  work  may  be  omitted,  rare 
being  taken  to  point  off  in  the  root  accordmg.  to  the  number  of  deci- 
mal periods  used. 

4.  The  pupil  will  acquire  greater  facility,  and  secure  greater  accura- 
cy,, by  keeping  units  of  like  order  under  each  other,  and  each  divisor 
opposite  tlie  corresponding  dividend,  by  ,the  use  of  the  lines,  as  shown 
in  the  operation. 

3.   What  is  the  square  root  of  576  ?  Ans.    24. 


SQUARE   ROOT. 

4.  What  is  the  square  root  of  6561  ? 

5.  What  is  the  square  root  of  444889  ? 

6.  What  is  the  square  root  of  994009  ? 

7.  What  is  the  square  root  of  29855296  ? 

8.  What  is  the  square  root  of  3486784401  ?    Ans.  59049. 

9.  What  is  the  square  root  of  54819198225  ? 

Note.  The  cipher  in  the  trial  rlivisor  may  be  omitted,  and  its  place, 
after  division,  occupied  by  the  trial  root  figui-e,  thus  forming  in  suc- 
cession only  complete  divisors. 


309 

An. 

•.    81. 

Ans. 

667. 

Ans. 

997. 

Ans. 

5464. 

10.    What  is  the  square 

root  of  2  ? 

2.           1.4142 +,  Ans. 

1 

100 

24 

96 

400 

281 

281 

11900 

2824 

11296 

V 

60400 

28282 

56564 

11.    Extract  the  square  roots  of  the  following  numbers; 


V3  =  1.7320508  + 
V5  =z  2.2360679  4- 
V6  irr  2.4494897  4- 


V7  =  2.6457513 -j- 
V8  =  2.82842714- 
VIO  =  3.1622776  4- 


'  12.   What  is  the  square  root  of  .00008836  ?       Ans.    .0094. 

13.  What  is  the  square  root  of  .0043046721  ?  Ans.  .06561. 

Notes.  5.  The  square  root  of  a  common  fraction  may  be  obtained 
by  extracting  the  square  roots  of  the  numerator  and  denominator 
separately,  provided  the  terms  are  perfect  squares ;  otherwise,  the 
fraction  may  first  be  reduced  to  a  decimal. 

6.  IMixed  numbers  may  be  reduced  to  the  decimal  form  before  ex- 
tracting the  root ;  or,  if  the  denominator  of  the  fraction  is  a  perfect 
square,  to  an  improper  fraction. 

14.  Extract  the  square  root  of  ^Wy  »         -^"^-   ti- 

15.  Extract  the  square  root  of  l^g'fl-  Ans.    ^. 

16.  Extract  the  square  root  of  §.  Ans.    .816496  +. 

17.  Extract  the  square  root  of  17f.  Ans.    4.1683  +• 


310 


EVOLUTION. 


APPLICATIONS. 

4J2!S.  An  Angle  is  the  opening  between  two  lines  tliat 
meet  each  otlier ;  thus,  the  two  lines,  A  B  and  A  C,  meeting, 
form  an  angle  at  A. 

■4SS.  A  Triangle  is  a  figure  having  three 
sides  and  three  angles,  as  A,  B,  C. 

4^4.    A  Eight- Angled  Triangle  is  a  tri- 
angle having  one  right  angle,  as  at  C. 

^2^5,    The  Ease  is  the  side  on  which  it 
stands,  as  A,  C. 

4:S©,    The  Perpendicular  is   the   side 
forming  a  right  angle  with  the  base,  as  B,  C. 

4:^7,  The  Hypotenuse  is  the  side  opposite  the  right  angle, 
as  A,  B. 

4:^8,  Those  examples  given  below,  which  relate  to  trian- 
gles and  circles,  may  be  solved  by  the  use  of  the  two  following 
principles,  which  are  demonstrated  in  geometry. 

1st.  Tlie  square  of  tlie  hypotenuse  of  a  right-angled  triangle 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides. 

2d.  The  areas  of  two  circles  are  to  each  other  as  the  squares 
of  their  radii,  diameters,  or  circumferences. 

1.  The  two  sides  of  a  right-angled  triangle  are  3  and  1 
feet ;  what  is  the  length  of  the  hypotenuse  ? 

Analysis.     Squarine 

OPERATION.  the  two  sides  and  add- 

32  =    9,  square  of  one  side.  i"?:.  "e  find  the  sum  to 

42  =  1 G,  sc.uare  of  the  other  side-       ^^  -^  '  '''''^  '^i"^^^  ^^"^  ^^"" 
—  IS  equal  to  the  square  of 

25,  square  of  hypotenuse.  the  hypotenuse,  we  ex- 

/25 5      A„g  tract  the  square  root,  and 

'  obtain  o  feet,  the  hypot- 

enuse.    Hence, 

To  find  tlie  hypotenuse.  Add  the  squares  of  the  tioo  sides, 
and  extract  the  square  root  of  the  sum. 

To  find  cither  of  the  shorter  sides.  Subtract  the  square  of 
the  given  side  from  the  square  of  the  hi/poteniise,  and  extract  the 
square  root  of  the  remainder. 


SQUARE  ROOT.  311 

EXAMPLES    FOR    TRACTICE. 

2.  If  an  army  of  55225  men  be  drawn  up  in  the  form  of  a 
square,  how  many  men  will  there  be  on  a  side  ?    Ans.   235. 

3.  A  man  has  200  yards  of  carpeting  l^^  yards  wide  ;  what 
is  the  lengtli  of  one  side  of  the  square  room  which  this  carpet 
will  cover  ?  Ans.    45  feet. 

4.  How  many  rods  of  fence  will  be  required  to  inclose  10 
acres  of  land  in  the  form  of  a  square  ?  A/ts.    1  GO  rods. 

5.  The  top  of  a  castle  is  45  yards  high,  and  the  castle  is  sur- 
rounded by  a  ditch  GO  yards  wide ;  required  the  length  of  a 
rope  that  will  reach  from  the  outside  of  the  ditcli  to  the  top 
of  the  castle.  Ans.    75  j-ards. 

G.  Required  the  height  of  a  May-pole,  which  being  broken 
39  feet  from  the  top,  the  end  struck  the  ground  15  feet  from 
the  foot.  Ans.    75  feet. 

7.  A  ladder  40  feet  long  is  so  placed  in  a  street,  that 
without  being  moved  at  the  foot,  it  will  reach  a  window  on 
one  side  33  feet,  and  on  the  other  side  21  feet,  from  the 
ground  ;  ■wliat  is  the  breadth  of  the  street  ?     Ans.   56.64  -\-  ft. 

8.  A  ladder  52  feet  long  stands  close  against  the  side  of  a 
building  ;  how  many  feet  must  it  be  drawn  out  at  the  bottom, 
that  the  top  may  be  lowered  4  feet  ?  Ans.    20  feet. 

9.  Tw'O  men  start  from   one   corner   of  a  park  one  mile 
square,  and  travel  at  the   same  rate.     A  goes  by  the  walk  ^ 
around  the  park,  and  B  takes  the  diagonal  path  to  the  opposite 
corner,  and  turns  to  meet  A  at  the  side.     How  many  rods 
from  the  corner  will  the  meeting  take  place  ?    Ans.  93.7  -\-  rods. 

10.  A  room  is  20  feet  long,  IG  feet  wide,  and  12  feet  high  ; 
what  is  the  distance  from  one  of  the  lower  corners  to  the  op- 
posite upper  corner  ?  Ans.    28.284271 -(-feet. 

11.  It  requires  63.39  rods  of  fence  to  inclose  a  circular 
field  of  2  acres ;  what  length  will  be  required  to  inclose  3 
acres  in  circular  form  ?  Ans.    77.63  +  rods. 

12.  The  radius  of  a  certain  circle  is  5  feet;  what  will  be 
the  radius  of  another  circle  containing  twice  the  area  of  the 
first?  A)is.    7.07106-1- feet. 


312  ■  EVOLUTION. 


CUBE  ROOT, 

4:29.  The  Cube  Root  of  a  number  is  one  of  the  three 
equal  factors  that  produce  the  number.  Thus,  the  cube  root 
of  27  is  3,  since  3  X  3  X  3  =  27. 

4;3©.  In  extracting  the  cube  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  cube  and  its 
root.  The  law  governing  this  relation  is  exhibited  in  the  fol- 
lowing examples  :  — 

Roots.  Cubes.  Roots.  Cubes. 

1111 

9  729  10  1,000 

99  907,299  100  1,000,000 

999         997,002,999  1000         1,000,000,000 

From  these  examples,  we  perceive, 

1st.  That  a  root  consisting  of  1  place  may  have  from  1  to 
3  places  in  the  cube. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root 
adds  three  places  to  the  cube.     Hence, 

If  we  pomt  off  a  number  into  three-JTgi(.re  periods,  com- 
mencing at  the  right  hand,  the  number  of  full  periods  and  the 
left  hand  full  or  partial  period  ivill  indicate  the  number  of 
places  in  the  cube  root,  the  highest  period  answering  to  the 
highest  figure  of  the  root. 

4:S1 .  1.  What  is  the  length  of  one  side  of  a  cubical  block 
containing  413494  solid  inches  ? 

OPERATION — COMMENCED.       AN.4.LYSIS.      Since   the    block  is  a 
413494  I  74  cube,  its  side  -will  be   the  cube  root  of 

343  its  soHd  contents,  which  \\c  will  pro- 

ceed  to   compute.     Pointing   off   the 

14/UU      /U4J4:  given  number,  the  two  periods   show 

that  there  will  be  two  figures,  tens  and 
units,  in  the  root.  The  tens  of  the  root  must  be  extracted  from  the 
first  period,  413  thousands.  The  greatest  cube  in  413  thousands  is 
343  thousands,  the  cube  of  7  tens  ;  we  therefore  write  7  tens  in  the 
root  at  the  right  of  the  given  number. 


CUBE   ROOT. 


313 


Since  the  entii-e  root  is  to  be  the  side  of  a  cube,  let  us  form  a 

cubical  block  (Fig.  I),  the  side 

of  which  is  70  inches  in  length. 

The  contents   of  this   cube  are 

70  X  70  X  70  =  343,000  solid 

inches,  which  Ave  subtract  from 

the  given  number.    This  is  done 

in  the  operation  by  subtracting 

the  cube  number,  343,  from  the 

first  period,  413,  and  to  the  re- 

^   mainder  bringing  down  the  sec- 

iW  cud  period,  making  the  entire 

PfjiiiiisSa^i^ca^^gsg^    I'emainder  70494. 

If  we  now  enlarge  our  culncal 
block,  (Fig.  I),  by  the  addition  of  70494  solid  inches,  in  sucli  a 
manner  as  to  preserve  the  cubical  form,  its  size  will  be  that  of 
the  required  block.  To  preserve  the  cubical  form,  the  addition 
must  be  made  upon  three  adjacent  sides  or  faces.  The  addition 
will  therefore  be  composed  of  3  flat  blocks  to  cover  the  3  faces, 
(Fig.  II)  ;  3  oblong  blocks  to  fill  the  vacancies  at  the  edges, 
(Fig.  Ill)  ;  and  1  small  cubical  block  to  fill  the  vacancy  at  the  cor- 
ner, (Fig.  IV).  Now,  the  thickness  of  this  enlargement  will  be  ihe 
additional  length  of  the  side  of  the  cube,  and,  consequently,  the 
second  figure  in  the  root.  To  find  thickness,  we  may  divide  solid 
^'--  ^^-  contents    by  surface,   or  area. 

But  the  area  of  the  3  oblong 
blocks   and   little    cube   cannot 
be   found  till  the   thickness  of 
the  addition  be  determined,  be- 
cause their  common  breadth  is 
equal  to  this  thickness.    We  will 
therefore  find  the   area   of  the 
three  flat  blocks,  which  is  sufla- 
ciently  near  the  whole  area  to  bo 
used  as  a  trial  divisor.    As  these 
are   each  equal  in  length   and 
breadth  to  the  side  of  the  cube 
whose  faces  they  cover,  the  M'hole 
area  of  the  three  is  70  X  70  X 
3  =■  14700  square  inches.     This  number  is  obtained  in  the  operation 
by  annexing  2  ciphers  to  three  times  the  square  of  7  ;  the  result 
being  written  at  the  left  hand  of  the  dividend.     Dividing,  we  obtain 
R.P.  14 


314 


EVOLUTION. 


Fis.  Ill 


OPERATION  —  CONTINtTED. 
413494 


IT. 


LO 

343 


74 


4,  the  probable  thickness  of  the  addition,  and  second  figure  of  the 

root.  With  tliis  assumed  figiu-e, 
■\ve  will  complete  our  divisor  by 
adding  the  area  of  the  4  blocks, 
before  undetermined.  The  3  ob- 
long blocks  are  each  70  inches 
long  ;  and  the  Httle  cube,  being 
equal  in  each  of  its  dimensions 
to  the  thickness  of  the  addition, 
must  be  4  inches  long.  Hence, 
their  united  length  is  70  -|-  70 
-j-  70  -|-  4  =  214.  This  number 
is  obtained  in  the  operation  by 
multiplying  the  7  by  3,  and  an- 
nexing the  4  to  the  product,  the 
result  being:  written  in  column 


being 

I,  on  the  next  Hne  below  the 
trial  divisor.  Multiplying  214, 
the  length,  by  4,  the  common 
width,  we  obtain  856,  the  area  of 
the  four  blocks,  which  added  to 
14700,  the  trial  divisor,  makes 
1.3556,  the  complete  divisor  ;  and 
multiplying  this  by  4,  the  second 
figure  in  the  root,  and  subtract- 
ing the  product  from  the  divi- 
dend, M'e  obtain  a  remainder  o! 
0270  solid  inches.  "Willi  this  re- 
mainder, for  the  same  reason  as 
before,  we  must  proceed  to  make 
a  new  enlargement.  But  since 
we  have  already  two  figures  in 
the  root,  answering  to  the  two 
periods  of  the  given  number, 
the  next  figure  of  the  root  must 
be  a  decnnal ;  and  wc  therefore 
annex  to  the  remainder  a  period 
of  three  decimal  ciphers,  mak- 
ing 8270.000  for  a  new  dividend. 
The  trial  divisor  to  obtain  the 
thickness  of  tliis  second  cnlargc- 


214  856 


14700 
1555G 


70494 
62224 


8270.000 


ment,  or  the  next  figure  of  the  root,  will  be  the  area  of  three  new  flat 
blocks  to  cover  the  three"  sides  of  the  cube  already  formed ;  and  this 


CUBE   ROOT. 


315 


surface,  (Fig.  IV,)  is  composed  of  1  face  of  each  of  the  flat  blocks 
ah-cacly  used,  2  faces  of  each  of  the  oblong  blocks,  and  3  faces  of 
tlie  little  cube.  But  we  have  in  the  complete  divisor,  15556,  1 
llice  of  each  of  the  flat  blocks,  oblong  blocks,  and  little  cube  ; 
and  in  the  correction  of  the  trial  divisor,  856,  1  face  of  each  of 
the  oblong  blocks  and  of  tlie  little  cube ;  and  in  the  square  of 
the  last  root  figure,  16,  a  third  face  of  the  little  cube.  Hence,  16 
-{r  856  -)-  15556  =  16428,  the  significant  figures  of  the  new  trial 

divisor.    This 
OrERATION  —  CONTINUED. 


ir. 


413i94  I  74.5 
343    


214 


856 


14700 
15556 


70494 
62224 


222.5   111.25 


1642800 
16539.25 


8270.000 
8269.625 


.oii) 


number  is  ob- 
tained in  the 
operation  by- 
adding  the 
square  of  the 
last  root  fig- 
ure mentally, 
and  coml^iu- 
ing  units  of 
like        order, 

thus :  16,  6,  and  6  are  28,  and  we  write  the  unit  figure  in  the  new 
trial  divisor ;  then  2  to  carry,  and  5  and  5  are  12,  Szc.  We  annex 
2  ciphers  to  this  trial  divisor,  as  to  the  former,  and  dividing,  obtain 
5,  the  third  figure  in  the  root.  To  complete  the  second  trial  di- 
visor, after  the  manner  of  the  first,  the  correction  may  be  found  bv 
annexing  .5  to  3  times  the  former  figures,  74,  and  midtiplying  this 
number  by  .5.  But  as  we  have,  in  column  I,  3  times  7,  with  4 
annexed,  or  214,  we  need  only  multiply  the  last  figure,  4,  by  3, 
and  annex  .5,  making  222.5,  which  multiplied  by  .5  gives  111.25, 
the  correction  required.  Then  we  obtain  the  complete  divisor, 
16539.25,  the  product,  8269.625,  and  the  remainder,  .375,  in  the 
manner  shown  by  the  former  steps.  From  this  example  and  analysis 
we  deduce  the  following 

Rule.  I.  Point  off  the  giveii  number  into  periods  of  three 
frjurcs  each,  counting  from  units'  place  toward  the  left  and  right. 

II.  Find  the  greatest  cuhe  that  does  not  exceed  the  left  hand 
period,  and  write  its  root  for  the  frst  figure  in  the  required 
root;  sidjtract  the  cube  from  the  left  hand  period,  and  to  the 
remainder  bring  doivn  the  next  period  for  a  dividend. 

III.  At  the  left  of  the  dividend  lorite  three  times  the  square 
of  tlie  first  figure  of  the  root,  and  annex  two  ciphers,  for  a  trial 
divisor  ;  divide  the  dividend  by  the  trial  divisor,  and  write  the 
quotient  for  a  trial  fgure  in  the  root. 


316 


ETOLUTION, 


IV.  Annex  tlie  trial  figure  to  three  times  the  former  figure, 
and  write  the  result  in  a  column  marked  I,  one  line  helow  the 
trial  divisor  ;  multiply  this  term  by  the  trial  figure,  and  write 
the  product  on  the  same  line  in  a  colwnn  marked  II ;  add  this 
term  as  a  correction  to  the  trial  divisor,  and  the  result  icill  be 
the  complete  divisor. 

V.  Multiply  the  complete  divisor  by  the  trial  figure,  and 
subtract  the  product  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend. 

VL  Add  the  square  of  the  last  figure  of  the  root,  the  last 
term  in  column  II,  and  the  complete  divisor  together,  and  annex 
two  ciphers,  for  a  neic  trial  divisor;  with  which  obtain  an- 
other trial  figure  in  the  root. 

VII.  Multiply  the  unit  figure  of  the  last  term  i?i  column  I 
by  3,  and  annex  the  trial  figure  of  the  root  for  the  next  term 
of  column  I ;  midtiply  this  residt  by  the  trial  figure  of  the  root 
for  the  next  term  of  column  11 ;  add  this  term  to  the  trial 
divisor  for  a  complete  divisor,  with  which  proceed  as  before. 

Notes.  1.  If  at  any  time  the  product  be  greater  than  the  dividend, 
diminish  the  trial  iigure  of  the  root,  and  correct  the  erroneous  -work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the 
trial  divisor,  and  another  period  to  the  dividend ;  then  proceed  as  be- 
fore -with  column  I,  annexing  both  cipher  and  trial  figure. 

EXAMPLES    FOR    PRACTICE. 

1.   What  is  the  cube  root  of  79.112  ? 

OPERATION. 

79.112  I  4.2928  +  ,  ^ng. 
64. 


4800    15112 

122 

244 

5044    10088 

529200   5024000 

12G9 

11421 

540621   4865589 

55212300  ' 158411000 

12872 

25744 

55238044  110476088 

5526379200  47934912000 

128768 

1030144 

5527409344  44219274752 

3714637248  rem. 


CUBE   ROOT.  317 

2.  What  is  the  cube  root  of  84G04-j19  ?  Ans.    439. 

3.  AVhut  is  the  cube  root  of  2o57947G01  ?  Ans.  1331. 

4.  What  is  the  cube  root  of  109G3240788375  ?    Ans.  22215. 

5.  What  is  the  cube  root  of  27UG71777U3218989G  ? 

Ans.    64G8G6. 
G.   Yv^hat  is  the  cube  root  of  .091125  ?  Ans.    .45. 

7.  What  is  the  cube  root  of  .000529475129  ?      Ans.  .0809. 

8.  AViiat  is  the  approximate  cube  root  of  .008649  ? 

Ans.    .2052  +  . 
Extract  the  cube  roots  of  the  followinj:  numbers  :  — 


72  =  1.259921  + 
^S  =  1.442249+ 
4'4  —  1.587401  + 


-^5  =  1.709975+ 
4'G  —  1.817120+ 
^7  =  1.912931+ 


APPLICATIOXS    IN    CUBE    KOOT. 

1.  What  is  the  length  of  one  side  of  a  cistern  of  cubical 
form,  containing  1331  solid  feet?  Ans.    11  feet. 

2.  The  pedestal  of  a  certain  monument  is  a  square  block  of 
granite,  containing  37324§  solid  inches ;  what  is  the  length 
of  one  of  its  sides  ?  Ans.    6  feet. 

3.  A  cubical  box  contains  474552  solid  inches  ;  what  is 
tiie  area  of  one  of  its  sides  ?  Ans.    42^  sq.  ft. 

4.  How  much  paper  will  be  required  to  make  a  cubical 
box  which  shall  contain  §|-  of  a  solid  foot?     Ans.  f  of  a  yard, 

5.  A  man  wishes  to  make  a  bin  to  contain  125  bushels,  of 
equal  width  and  depth,  and  length  double  the  width  ;  what 
must  be  its  dimensions  ?  Ans.  W^idth  and  depth,  51.223  + 
inches;  length,  102.44G  +  inches. 

XoTE.  Spheres  are  to  each  other  as  the  cubes  of  theii-  diameters  or 
circumferences. 

G.  There  are  two  spheres  whose  solid  contents  are  to  each 
other  as  27  to  343  ;  what  is  the  ratio  of  their  diameters  ? 

Analysis.  Since  spheres  are  to  each  other  as  the  cubes  of  their 
diameters,  the  diameters  will  be  to  each  other  as  the  cube  roots  of 
the  spheres ;  and  ^27  =  3,  ^343  =  7 ;  hence  the  diameters  required 
are  as  3  to  7. 


318  ARITHMETICAL   PROGRESSION. 

7.  The  diameter  of  a  sphere  containing  1  solid  foot  is  14.9 
inches ;  Avhat  is  the  diameter  of  a  sphere  containing  2  solid 
feet?  Ans.   18.7  +  inches. 

8.  If  a  cable  4in.  in  circumference,  will  support  a  sphere  2ft. 
in  diameter,  what  is  the  diameter  of  that  sphere  which  will  be 
uipported  by  a  cable  5in.  in  circumference?        Ans.  2.32+ ft. 

ARITHMETICAL   PROGRESSION. 

4S2.  An  Arithmetical  Progression,  or  Series,  is  a  series 
of  numbers  increasing  or  decreasing  by  a  common  difference. 
Thus,  3,  5,  9,  11,  &c.,  is  an  arithmetical  progression  with  an 
ascending  series,  and  13,  10,  7,  4,  «&c.,  is  an  arithmetical  pro- 
gression with  a  descending  series. 

^JB3.  The  Terms  of  a  series  are  the  numbers  of  which  it 
is  composed. 

4JI-1.    The  Extremes  are  the  first  and  last  terms. 

4:»I«S.    The  Means  are  the  intermediate  terms. 

4S@.  The  Common  Difference  is  the  difference  between 
any  two  adjacent  terras. 

43^.  There  are  Jive  parts  in  an  arithmetical  series,  any 
three  of  which  being  given,  the  other  two  may  be  found. 
The-y  are  as  follows :  the  Jirst  term,  last  te)-ni,  common  differ' 
ence,  numljer  of  terms,  and  sum  of  all  the  terms. 

CASE    I. 

4:38.  To  find  the  last  term  when  the  first  term, 
common  diflbrcncc,  and  number  of  terms  are  given. 

Let  2  be  the  first  term  of  an  ascending  series,  and  3  the 
common  difference ;  then  the  series  will  be  written,  2,  5,  8,  11, 
14,  or  analyzed  thus :  2,  2-f3,  2-|-3  +  3,  2-|-3-|-3-f-3, 
2  -\-  3  -j-  ■'!  -J-  '•>  -j—  3. 

Here  we  see  that,  in  an  ascending  series,  we  obtain  the 
x"rnnd  term  I)y  adding  the  common  dilf(>rence  once  to  the  first 
tiiiii  ;  the  tliird  terra,  by  adding  the  conniion  difierence  twice 
to   the  first  term ;   and,  in  general,  we  obtain  any  term  by 


ARITHMETICAL   PROGRESSION.  319 

adding  the  common  difference  as  many  times  to  the  first  term 
as  there  ai'e  terms  less  one. 

Note.  The  analysis  for  a  descending  series  would  be  similar. 
Hence, 

Rule.  31idtiply  the  common  difference  hy  the  number  of 
terms  less  one,  and  add  the  product  to  the  Jirst  term,  if  the 
series  be  ascending,  and  subtract  it  if  the  series,  be  descending. 

EXAMPLES. 

1.  The  first  term  of  an  ascending  series  is  4,  the  common 
difference  3,  and*  the  number  of  terms  19 ;  what  is  the  last 
term  ?  A7is.    58. 

2.  What  is  the  13th  term  of  a  descending  series  Avhose  first 
term  is  75,  and  common  difference  5  ?  Ans.    15. 

3c  A  boj  bought  18  hens,  paying  2  cents  for  the  first,  5 
cents  for  the  second,  and  8  cents  for  the  third,  in  arithmetical 
progression  ;  what  did  he  pay  for  the  last  hen  ? 

4.  What  is  the  40th  term  of  the  series  ^,  -|,  1,  1^,  &c.  ? 

Ans.    10^-. 

5.  A  man  travels  9  days;  the  first  day  he  goes  20  miles, 
the  second  25  miles,  increasing  5  miles  each  day;  how  far 
does  he  travel  the  last  day  of  his  journey  ?     Ans.  GO  miles. 

6.  What  is  the  amount  of  $100,  at  7  per  cent.,  for  45 
years  ?  $100  +  $7  X  45  =  $415,   Ans. 

CASE    II. 

4Si5.  To  find  the  common  diiference  when  the 
extremes  and  number  of  terms  are  given. 

Referring  to  the  series,  2,  5,  8,  11,  14,  analyzed  in  -^SS, 
we  readily  see  that,  by  subtracting  the  first  term  from  any 
term,  we  have  left  the  common  difference  taken  as  many  times 
as  there  are  terms  less  one ;  thus,  by  taking  away  2  in  the 
fifth  term,  2  -|-  3  -[-  3  -}-  3  -j-  3,  we  have  3  taken  4  times. 
Hence, 

Rule.  Divide  the  difference  of  the  extremes  by  the  number 
of  terms  less  one. 


320  ARITHMETICAL  PROGRESSION. 

EXAMPLES. 

1.  The  first  term  is  2,  the  last  term  is  17,  and  the  number 
of  terms  is  6  ;  what  is  the  common  difference  ?  Aiis.    3. 

2.  A  man  has  seven  children,  whose  ages  are  in  arithmetical 
progression;  the  youngest  is  2  years  old,  and  the  eldest  14; 
what  is  the  cammon  ditference  of  their  ages  ?   A7is.  2  years. 

3.  The  extremes  of  an  arithmetical  series  are  1  and  50^, 
and  the  number  of  terms  is  34 ;  what  is  the  common  difference  ? 

4.  An  invalid  commenced  to  walk  for  exercise,  increasing 
the  distance  daily  by  a  conuuon  difference^  the  first  day  he 
walked  3  miles,  and  the  14th  day  9^  miles;  how  many  miles 
did  he  walk  each  day  ? 

Note.  "When  we  have  found  the  common  difference  we  may  add  it 
once,  twice,  Sec,  to  the  fii-st  term,  and  we  have  tl\e  series,  and  conse- 
quently the  means. 

Ajis.   3,  3i,  4,  A^,  5,  5i,  &c. 

CASE    III. 

410.  To  find  the  number  of  terms  when  the  ex- 
tremes and  common  difference  are  given. 

Examining  the  series,  2,  5,  8,  11,  14,  analyzed  in  4^38, 
we  also  see  that  after  taking  away  the  jrrst  term  irom  any 
term,  we  have  left  the  common  difference  taken  as  many 
times  as  the  number  of  terms,  less  1.     Hence, 

Rule.  Divide  the  difference  of  the  extremes  ly  the  common 
difference,  and  add  1  to  the  quotient. 

EXAMPLES. 

1.  The  extremes  are  7  and  43,  and  the  common  difference 
is  4  ;  what  is  the  number  of  terms  ?  Aiis.    10. 

2.  The  first  term  is  2^,  the  last  term  is  40,  and  the  common 
difference  is  7^ ;  what  is  the  number  of  terms  ?         Ans.    G. 

3.  A  laborer  agreed  to  build  a  fence  on  the  following  con- 
ditions :  for  the  first  rod  he  was  to  have  G  cents,  with  an 
increase  of  4  cents  on  each  successive  rod ;  the  last  rod  came 
to  22G  cents;  how  many  rods  did  he  build  ?    Ans.   5G  i*ods. 


GEOMETRICAL   PROGRESSION.  321 

CASE    IV. 

441.  To  find  the  sum  of  all  the  terms  when  the 
extremes  and  number  of  terms  ^.re  given. 

To  deduce  a  rule  for  finding  the  swji  of  all  the  terms,  we 
will  take  the  series  2,  5,  8,  11,  14,  writing  it  under  itself  in  an 
inverse  order,  and  add  each  term ;  thus, 

9  _j_    5  4-    3  +  1 1  -f  14  =  40,  once  the  sura. 
14-1-11+    8 -]-    5 -f    2  =  40,     "       "       « 
lG-f-lG-]-lG-[-lG-fl6==:80,  twice  the  sum. 

Here  we  perceive  that  IG,  the  sum  of  the  extremes,  multi- 
plied by  5,  the  number  of  terms,  equals  80,  which  is  tivice  the 
sum  of  the  series.  Dividing  80  by  2  gives  40,  which  is  the 
sum  required.     Kence, 

Rule.  MuMlphj  the  sum  of  tlie  extremes  hy  the  niimher  gJ 
terms,  and  divide  the  'product  hy  2. 

EXA:\irLES. 

1.  The  extremes  are  5  and  32,  and  the  number  of  terms  12  ; 
what  is  the  sum  of  all  the  terras  ?  Ans.    '2'22. 

2.  How  many  strokes  does  a  common  clock  make  in  12 
hours  ?  Ans.    78  strokes. 

3.  What  debt  can  be  discharged  in  a  year  by  weekly  pay- 
ments in  arithmetical  progression,  the  first  being  $24,  and  the 
last  $1224?  J^ns.    $32448. 

4.  Suppose  100  apples  were  placed  in  a  line  2  yards  apart, 
and  a  basket  2  yards  from  the  first  apple ;  how  far  would  a 
boy  travel  to  gather  them  up  singly,  and  return  with  each 
separately  to  the  basket  ?  Ans.   20200  yards. 

GEOMETRICAL  PROGRESSION. 

44®.  A  GeometricpJ  Progression  is  a  series  of  numbere 
inci'easing  or  decreasing  by  a  constant  multiplier. 

When  the  multiplier  is  greater  than  a  unit,  the  senes  is 

14* 


322  GEOMETRICAL  PEOGRESSION. 

ascending;  thus,  2,  6,  18,  54,  1G2,  is  an  ascending  series,  in 
which  3  is  the  rauUiplier. 

Wlaen  the  nudtiplier  is  less  than  a  unit,  the  series  is  descend- 
ing; thus,  1G2,  54,  18,  6,  2,  is  a  descending  series,  in  which  \ 
is  the  muhipHer. 

^143.    The  Katio  is  the  constant  rauhiplier. 

4l:'14.  In  every  geometrical  progression  there  are  five 
parts  to  be  considered,  any  tliree  of  wliich  being  given,  the 
other  hvo  may  be  determined.  Tliey  are  as  follows:  T\\c Jirst 
term,  last  term,  ratio,  number  of  terms,  and  the  sum  of  all  tJie 
terms. 

The  first  and  last  terms  are  the  extremes,  and  the  intex'me- 
dlate  terms  are  the  means. 

CASE    I. 

44i5.  To  find  any  term,  the  first  term,  the  ratio, 
and  number  of  terms  being  given. 

Tlie  first  term  is  supposed  to  exist  independently  of  the 
ratio.  Using  the  ratio  once  as  a  factor,  we  have  the  second 
term  ;  using  it  twice,  or  its  second  power,  we  have  the  third 
term  ;  using  it  three  times,  or  its  third  power,  we  have  the 
fourth  term ;  and,  in  general,  the  power  of  the  ratio  in  any 
term  is  one  less  than  the  number  of  the  term.  The  ascending 
series,  2,  6,  18,  54,  may  be  analyzed  thus:  2,  2  X  ">,  2  X 
3X3,  2X3X3X3. 

In  this  illustration  we  see  that 

1st  term,  2,  is  independent  of  the  ratio. 

2d  "  6  =  2X3  =  the  first  term  into  the  1st  power  of 
ihfi  ratio. 

3d  term,  18  =  2  X  3^  1=  the  first  term  into  the  2d  power 
of  the  ratio. 

4th  term,  54  =:  2  X  3  ^  =  the  first  term  into  the  3d  power 
of  the  ratio.     Hence 

Rule,  ^ftdtipli/  the  first  term  hj  that  power  of  the  ratio 
denoted  hy  the  number  of  terms  less  1. 


GEOMETRICAL   rROGllESSION.  323 

EXAMPLES. 

1.  The  first  term  of  a  geometrical  series  is  4,  the  ratio  is  3 ; 
what  is  the  9th  term  ?  Ans.    4  X  S^  zrr  2G244. 

2.  The  first  term  is  1024,  the  ratio  J,  and  the  number  of 
terms  8;  what  is  the  hist  term  ?  Ans.   yg^. 

3.  A  boy  bought  9  oranges,  agreeing  to  pay  1  mill  for  the 
first  orange,  2  mills  for  the  second,  and  so  on ;  what  did  the 
last  orange  cost  him  ?  Ans.    $.256. 

4.  The  first  term  is  7,  the  ratio  j-,  and  the  number  of  terms 
7  ;  what  is  the  last  term  ?  Ans.    iq^xjj' 

5.  What  is  the  amount  of  $1  at  compound  interest  for  5 
years,  at  7  per  cent,  per  annum  ?  Ans.    $1.40255  -[-• 

Note.  In  the  above  example  the  first  term  is  $1,  the  ratio  is  $1.07, 
and  the  number  of  terms  is  6. 

6.  A  drover  bought  7  oxen,  agreeing  to  pay  $3  for  the  first 
ox,  $9  for  the  second,  $27  for  the  third,  and  so  on ;  what  did 
the  last  ox  cost  him  ?  Ans.    $2187. 

CASE    II. 

4411.  To  find  the  sum  of  all  the  terms,  the  ex- 
tremes and  ratio  being  given. 

If  Ave  take  the  series  2,  8,  32,  128,  512,  in  which  the  ratio 
is  4,  multiply  each  term  by  the  ratio,  and  add  the  terms  thus 
multiplied,  we  shall  have 

8  +  32  +  128  +  512  +  2048  =.  2728  ={,^-  [Irten'J"" 

But  2  +  8  +  32  +  128  +  512z:r      682  =  j  ^  telt!""  "'  '''" 

Hence,  by  subtracting,  we  get  2048  —  2  —  204G  =  {  o/lnthTtmaJ'^'^ 
Dividing  by  3,  the  ratio  less  one,  204G  -|-  3  =  682  =r  |  ji .""emc.^"'"  " 

The  subtraction  is  performed  by  taking  the  lower  line  or 
series  from  the  upper.  All  the  terms  cancel  except  2048  and 
2.  Taking  their  difference,  which  is  3  times  the  sum,  and  di- 
viding by  3,  the  ratio  less  one,  we  must  have  the  sum  of  all 
the  terms.    Hence 


o2-i  PROMISCUOUS   EXAMPLES. 

Rule.    JlluUipIt/  the  greater  extreme  hy  the  ratio,  subtract 

the  less  extreme  from  the  product,  and  divide  the  remainder 

by  the  ratio  less  1. 

Note.  Let  every  decreasing  series  be  inverted,  and  the  first  term 
called  the  last ;  then  the  ratio  will  be  greater  than  a  luiit.  If  the  series 
be  infinite,  the  first  term  is  a  cipher. 

EXAMPLES. 

1.  The  first  terra  is  2,  the  last  term  512,  and  the  ratio  3 ; 
■u'liat  is  the  sum  of  all  the  terms?  Ans.    707. 

2.  The  first  term  is  4,  the  last  term  is  2G2144,  and  the 
ratio  is  4;  what  is  the  sum  of  the  series?       Ans.    349524. 

3.  The  first  term  of  a  descending  series  is  1G2,  the  last 
term  2,  and  the  ratio  \  ;  what  is  the  sum  ?  Ans.    242. 

4.  "What  is  -the  value  of  \,  ^-^,  y^,  &c.,  to  infinity  ?  Ans.  \. 

I^^OTE.  In  the  following  examples  v.'c  first  find  the  last  term  by  the 
Rule  under  Case  I. 

5.  What  yearly  debt  can  be  discharged  by  monthly  pay- 
ments, the  first  being  $2,  the  second  $6,  and  the  third  $18, 
and  so  on,  in  geometrical  progression  ?  Ans.    $531440. 

6.  If  a  grain  of  wheat  produce  7  grains,  and  these  be  sown 
the  second  year,  each  yielding  the  same  increase,Jiow  mar.y 
bushels  Avill  be  produced  at  this  rate  in  12  years,  if  1000  grains 
make  a  pint?  Ans.  252315  bu.  4i  qt. 

7.  Six  persons  of  the  Morse  family  came  to  this  country 
200  years  ago;  suppose  that  their  number  has  doubled  every 
20  years  since,  what  would  be  their  number  now  ? 

Note.  The  other  cases  in  Progression  will  be  found  in  tlio  Higher 
Arithmetic. 


PROMISCUOUS  EXAMPLES. 

\.   One  half  tho  sum  of  two  numbers  is  800,  and  one  lialf  the 
difference  of  the  same  numbers  is  200  ;  what  are  the  numbers  ? 

Am.    1000  and  600. 

2.  What  number  is  that  to  which,  if  you  add  f  of  {^^  of  itseLP, 
the  sum  will  be  01  ?  Ajis.   55. 

3.  What  part  of  a  day  is  3  h.  21  min.  15  sec.  ?  Ans.    ^-[^^. 


PROMISCUOUS   EXAMPLES.  S25 

4.  A  commission  merchant  I'eceived  70  bags  of  wheat,  each  con- 
taining 3  bu.  3  pk.  o  qt. ;  how  many  bushels  did  he  receive  ? 

5.  Four  men,  A,  B,  C,  and  D,  are  in  possession  of  $1100;  A 
has  a  certain  sum,  B  has  twice  as  much  as  A,  C  has  $300,  and  D 
has  $200  more  than  C  ;  how  many  dollars  has  A  ?       Ans.    $100. 

6.  At  a  certain  election,  3000  votes  were  cast  for  three  candi- 
dates, A,  B,  and  C  ;  B  had  200  more  votes  than  A,  and  C  had  800 
more  than  B  ;  how  manv  votes  were  cast  for  A  ?  Ans.   GOO. 

7.  What  part  of  17^  IS  31?  Ans.   f|. 

8.  Tlie  difference  between  4  and  -I  of  a  number  is  10 ;  what  is 
the  number  ?  Ans.   560. 

9.  A  merchant  bought  a  hogshead  of  rum  for  $28.3<5  ;  how 
much  v.'ater  must  be  added  to  reduce  the  first  cost  to  35  cents  per 
gallon  ?  Ans.    18  gal. 

10.  A  and  B  traded  with  equal  simis  of  money  ;  A  gained  a  sum 
equal  to  l  of  his  stock ;  B  lost  $200,  and  then  he  had  ^  as  much  as 
A  ;  liow  much  was  the  original  stock  of  each  ?  Ans.    $500. 

11.  A  farmer  sold  17  bushels  of  barlej-,  and  13  bushels  of  wheat, 
for  $31.55  ;  he  received  for  the  wheat  35  cents  a  bushel  more  than 
for  the  barley  ;  what  was  the  price  of  each  per  bushel  ? 

A71S.    Barley,  $.90;  wheat,  $1.25. 

12.  What  is  the  interval  of  time  between  JIarch  20,  21  minutes 
past  3  o'clock,  P.  M.,  and  April  11th,  5  minutes  past  7  o'clock, 
A.  M.  ?  Ans.    21  da.  15  h.  44  min. 

13.  What  o'clock  is  it  when  the  time  from  noon  is  ^^^  of  the . 
time  to  midnight  ?  Ans.    5  o'clock  24  min.  P.  M. 

14.  What  is  the  least  number  of  gallons  of  witie  that  can  be  ship- 
ped in  either  hogsheads,  tierces,  or  barrels,  just  filhng  the  vessels, 
without  deficit  or  excess  ?  Ans.    126  gal. 

15.  A  ferryman  has  four  boats  ;  one  v>-ill  carry  8  barrels,  another 
9,  another  15,  and  another  16  ;  V)hat  is  the  smallest  number  of  bar- 
rels that  will  make  full  freight  for  any  one,  and  all  of  the  boats  ? 

16.  A  and  B  have  the  same  income;  A  saves  |-  of  his,  but  B, 
by  spending  $30  a  year  more  than  A,  at  the  end  of  four  years  finds 
himself  $40  in  debt ;  what  is  their  income,  and  how  much  does 
each  spend  a  year  .f*  C  Income,     $160. 

A71S.  <  A  spends  $140. 
(  B  spends  $170. 

17.  If  a  load  of  plaster  weighing  1825  pounds  cost  $2.19,  how 
much  is  that  per  ton  of  2000  pounds  ?  Ans.    $2.40. 

18.  If  21  yards  of  cloth  If  yards  wide  cost  $3.37-|,  what  will  be 
the  cost  of  36-1  yards  l-l  yards  wide  ?  Ans.    $52. 779. 

19.  I  lend  my  neighbor  $200  for  6  months  ;  how^  long  ought  he 
to  lend  me  $1000  to  balance  the  favor  ? 

20.  Bought  railroad  stock  to  the  amount  of  $2356.80,  and  found 
that  the  sum  invested  was  40  per  cent  of  what  I  had  left ;  what 
sum  had  I  at  first  ?  Ans.    $8248.80. 

21.  20  per  cent,  of  f  of  a  number  is  what  per  cent,  of  #  of  it  ? 

Arcs.    121 


325  PROMISCUOUS   EXAMPLES. 

22.  Divide  a  prize  of  $10200  among  60  privates,  6  subaltern 
oificers,  3  lieutenants,  and  a  commander,  giving  to  each  subaltern 
double  the  share  of  a  private,  each  lieutenant  3  times  as  much  as 
the  subaltern,  and  to  the  commander  double  that  of  a  lieutenant ; 
how  much  is  each  man's  share?    Ans.  Com.  $1200;  eacli  man,  $100. 

23.  A  is  51  miles  in  advance  of  B,  who  is  in  pursuit  oi  him  ;  A 
travels  16  miles  per  hour,  and  B  19  ;  in  how  many  hours  \\  ill  B 
overtake  A  ? 

24.  How  much  wool,  at  20,  30,  and  54  cents  per  pound,  must  be 
mixed  with  95  pounds  at  50  cents,  to  make  the  whole  mixture 
worth  40  cents  per  pound  ? 

Am.    133  lb.  at  20  ;  95  lb.  at  30  ;  190  lb.  at  54  cents. 

25.  If  240  bushels  of  wheat  are  purchased  at  the  rate  of  18 
bushels  for  $22^,  and  sold  at  the  rate  of  22^  bushels  for  $33.75, 
M'hat  is  the  profit  on  the  whole  ?  Ans.    $60. 

26.  'My  horse,  wagon,  and  harness  together  are  worth  $169  ;  the 
wagon  is  worth  4  times  the  harness,  and  the  horse  is  worth  double 
the  wagon  ,  what  is  the  value  of  each  ?  C  Horse,      $104. 

Ans.  <  Wagon,   $  52. 
(  Harness,  $  13. 

27.  The  shadow  of  a  tree  measures  42  feet ;  a  staff'  40  inches  in 
length  casts  a  shadow  18  inches  at  the  same  time  ;  what  is  the 
height  of  the  tree  ?  Ans.  93^  ft. 

28.  If  a  piece  of  land  40  rods  long  and  4  rods  wide  make  an 
acre,  how  wide  must  it  be  to  contain  the  same  if  it  be  but  25  rods 
long  P  A}is.    6|  rods. 

29.  A,  B.  and  C  are  employed  to  do  a  piece  of  work  for  $26.45 ; 
A  and  B  together  are  supposed  to  do  -f  of  it,  A  and  C  -^^,  and  B 
and  C  ^f^,  and  paid  proportionally  ;  how  much  must  each  receive  ? 

30.  If  12  ounces  of  wool  make  2^  yards  of  cloth  that  is  6  quar- 
ters wide,  how  many  pounds  of  wool  will  it  take  for  150  yards  of 
cloth  4  quarters  wide  ? 

31.  Six  persons.  A,  B,  C,  D,  E,  and  F,  are  to  share  among  them 
$6300  :  A  is  to  have  i  of  it,  B^,  C  |,  ])  is  to  have  as  much  as  A 
and  C  together,  and  the  remainder  is  to  be  divided  between  E  and 
F  in  the  proportion  of  3  to  5  ;  how  much  does  each  one  receive  ? 

32.  AVhat  is  the  amount  of  $200  for  8  years  at  6  per  cent,  com- 
pound interest  ?  Ans.    $318,769. 

33.  A  garrison,  consisting  of  360  men,  was  provisioned  for  6 
months ;  but  at  the  end  of  5  months  they  dismissed  so  many  of  the 
men  that  the  remaining  provision  lasted  5  months  longer ;  how 
many  men  were  sent  away  P 

34.  A  certain  principal,  at  compound  interest  for  5  years,  at  6 
per  cent.,  •will  amount  to  $015;). 113  ;  in  wiiat  time  will  tli-o  same 
j)rinciiial  amount  to  the  same  sum,  at  6  per  cent,  simple  interest? 

Ans.  5  yr.  7  mo.  lO.S-f-da. 

35.  Paid  $148,352  for  0728  feet  of  jiine  hunber ;  how  much  waa 
that  per  tliousand  ? 


I 


PROxAIISCUOUS   EXAMPLES.  327 

3G.  Comparlns^  two  luimbers,  4S0  was  found  to  he  their  least 
common  multiple,  and  123  their  greatest  common  divisor ;  what 
is  the  product  of  the  numbers  compared  ?  Ans.  11,10S). 

37.  Eight  workmen,  hiboring  7  hours  a  day  for  15  days,  were 
able  to  execute  -^  of  a  job  ;  in  how  many  days  can  they  complete  the 
residue,  by  working  9  hours  a  day,  if  4  workmen  are  added  to  their 
number?  Ans.    lo|  days. 

38.  If  a  hall  36  feet  long  and  9  feet  wide  require  3li  yards  of 
carpeting  1  yard  wide  to  cover  the  iioor,  how  many  yards  \\  yards 
wide  will  cover  a  floor  GO  feet  long  and  21  feet  wide  ? 

Alls.  144  yards. 

39.  A,  B,  and  C  traded  in  company  ;  A  put  in  $1  as  often  as  B 
put  in  $3,  and  B  put  in  $2  as  often  as  C  put  in  $<3  ;  B's  money  was 
in  twice  as  long  as  ("s,  and  A's  twice  as  long  as  B's  ;  they  gained 
$d2.50  ;  how  much  was  each  man's  share  of  the  gain  ?  C  A's,  $12. 

.4/t5. -{B's,  $18. 
(C's,   $22.50. 

40.  A  and  B  found  a  watch  worth  $45,  and  agreed  to  divide  the 
value  of  it  in  the  ratio  of  |-  to  -| ;  how  much  was  each  one's  share  ? 

,       S  $20,  A's. 
^  $2o,  B's. 

41.  A  man  received  $33.25  interest  on  a  sum  of  money,  loaned 
5  years  preA'ious,  at  7  per  cent. ;  what  was  the  sum  lent  ? 

Ans.   $95. 

42.  The  diameter  of  a  ball  weighing  32  pounds  is  6  inches  ; 
what  is  the  diameter  of  a  ball  weighing  4  pounds  ?     A^is.    3  inches. 

43.  Divide  $360  in  the  proportion  of  2,  3,  and  4. 

Ans.  |80,  $120,  $160. 

44.  If  by  working  6f  hours  a  day  a  man  can  accomplish  a  job  in 
121  days,  how  many  days  will  be  required  if  he  work  Si  hours  per 
day?  Ans.    9^^^  days. 

45.  An  open  court  contains  40  square  yards ;  how  many  stones, 
9  inches  square,  will  be  required  to  pave  it  ?  Ans.    640. 

46.  A  drover  paid  $76  for  calves  and  sheep,  paying  $3  for  calves, 
and  $2  for  sheep  ;  he  sold  1  of  his  calves  and  |-  of  his  sheep  for 
$23,  and  in  so  doing  lost  8  per  cent,  on  their  cost ;  how  many  of 
each  did  he  purchase  ?  Ans.    12  calves  ;  20  sheep. 

47.  If  a  cistern,  17-^  feet  long,  lO-J-  ])road,  and  13  deep,  hold  548 
barrels,  how  many  barrels  will  that  cistern  hold  that  is  16  feet  long, 
7  broad,  and  15  deep  ?  Ans.    384  bbls. 

48.  If  12  men,  working  9  hours  a  day,  for  15|-  days,  were  able  to 
execute  f  of  a  job,  how  many  men  may  be  withdrawn,  and  the  resi- 
due be  finished  in  15  days  more,  if  the  laborers  are  employed  only 
7  hours  a  day  ?  Ans.    4  men. 

49.  A  general  formed  his  men  into  a  square,  that  is,  an  equal 
number  in  rank  and  file,  and  found  that  he  had  59  men  over ;  and 
increasing  the  number  in  both  rank  and  file  by  1  man,  he  wanted  84 
men  to  complete  the  square ;  how  many  men  had  he  ?   Ans.  5100. 


823  PROMISCUOUS   EXAEiriES. 

50.  Bought  wheat  at  $1.50  per  bushel,  corn  at  $.75  per  bushel, 
and  barley  at  $.60  per  bushel  ;  the  wheat  cost  twice  as  much  as  the 
corn,  and  the  corn  twice  as  much  as  the  barley  ;  of  the  sum  paid, 
§243  and  -|-  of  the  whole  was  for  wheat,  and  $153  and  ^V  of  the 
whole  Avas  for  the  corn ;  how  many  bushels  of  grain  did  I  purchase  ? 

Ans.    756. 

51.  Divide  $630  among  3  persons,  so  that  the  second  shall  have 
■|  as  much  as  the  first,  and  the  third  -^  as  much  as  the  other  two ; 
what  is  the  share  of  each  ?  CI  st,  $240. 

,     Ans.VlA,  $180. 
^3d,  $210. 

52.  Bought  a  hogshead  of  molasses  for  $28,  and  7  gallons 
leaked  cut ;  at  what  rate  per  gallon  must  the  remainder  be  sold  to 
gain  20  per  cent.  ? 

53.  20  per  cent,  of  |-  of  a  number  is  how  many  per  cent,  of  2 
times  -|-  of  11  times  the  number  ?  Ans.    1^. 

54.  B  and  C,  trading  together,  find  their  stock  to  be  worth  $3o00, 
of  which  C  owns  $2100 ;  they  have  gained  40  per  cent,  on  their  first 
capital ;  what  did  each  put  in  ?  .       5  ^>  $1000. 

^^^^-  \  C,  $1500. 

55.  If  the  ridge  of  a  building  be  8  feet  above  the  beams,  and  the 
building  be  32  feet  wide,  what  must  be  the  length  of  rafters  ? 

56.  If  12  workmen,  in  12  days,  working  12  hours  a  day,  can 
make  up  75  yards  of  cloth,  :|  of  a  yard  wide,  into  articles  of  clothing : 
how  many  yards,  1  yard  wide,  can  he  made  up  into  like  articles,  by 
10  men,  working  9  days,  8  hours  each  day  ?  Ans.    23-j'^g. 

57-  A  grocer  sells  a  farmer  ICO  pounds  of  sugar,  at  12  cents  a 
pound,  and  makes  a  profit  of  9  per  cent. ;  tlie  former  sells  him  100 
pounds  of  beef,  at  6  cents  a  ])ound,  and  makes  a  profit  of  10  per 
cent. ;  who  gains  the  more  by  the  trade,  and  how  much  ? 

Ans.    The  grocer  gains  $.415+  more. 

58.  In  1  yr.  4  mo.  $311.50  amounted  to  $336.42,  at  simple 
interest ;  what  was  the  rate  per  cent.  ?  Ans.    G. 

59.  Three  persons  engage  to  do  a  piece  of  work  for  $20 ;  A  and 
B  estimate  that  they  do  4-  of  it,  A  and  C  that  they  do  |-  of  it,  and  B 
and  C  that  thcv  do  |  of  it ;  according  to  this  estimate,  what  jiart  of  the 
$20  should  each  man  receive  ?     Ans.   A's,  $1  If  ;  B's,  $55  ;  C's,  $2f 

60.  Paid  $375,  at  the  rate  of  2^  i)er  cent.,  for  insurance  on  a 
cotton  factory  and  the  machinery ;  for  what  amount  Avas  the  policy 
given  ? 

61.  A  merchant  bought  goods  in  Boston  to  the  amount  of  $1000, 
and  gave  his  note,  dated  Jan.  1,  1857,  on  interest  after  3  months; 
six  months  after  the  note  was  given  he  paid  $560,  and  5  months 
after  the  first  pavment  he  paid  $406  ;  Avhat  was  due  Aug  23,  1859  ? 

Ans.  $0G.63-(- 

62.  If  I  of  A's  money  be  equal  to  -|  of  B's,  and  ■§-  of  B's  be  equal 
to  ^  of  C's,  and  -?-  of  C's  ])e  equal  to  ^  of  D'e,  and  D  has  $45  more 
tlmn  C,  howjtnuch  has  each  ?  4       S  A,  8378  ;  C,  $360  ; 

^^"^-  I  B,  $336  ;  D,  $405. 


PROMISCUOUS   EXAMPLES.  329 

63.  A  owed  B  $900,  to  be  paid  in  3  years  ;  but  at  the  expiration 
of  9  mouths  A  agreed  to  pay  $300  if  B  would  wait  long  enough  for 
the  balance  to  compensate  for  the  advance ;  how  long  should  B 
wait  after  the  expiration  of  the  3  years  ?  Aiis.    13-|-  mo. 

C4.  A  certain  clerk  receives  $800  a  year ;  his  expenses  equal  J^ 
of  Mhat  he  saves ;  how  much  of  his  salary  does  he  save  yearly  P 

65.  A  merchant  sold  cloth  at  $1  per  yard,  and  made  10  per  cent, 
profit ;  what  would  have  been  iiis  gain  or  loss  had  he  sold  it  at  $.87^ 
per  ya.rd  ?  Aiis.    Loss,  3|-  per  cent. 

0J_9_ 

66.  What  is  the  cube  of  ^~  Ans.  U. 

29  J^  ^^ 

63 

67.  What  is  the  cube  root  of  — ^  A'ls.   f . 

68.  A  miller  is  required  to  grind  100  bushels  of  provender  worth 
50  cents  a  bushel,  from  oats  worth  20  cents,  corn  worth  35  cents, 
rye  worth  60  cents,  and'  wheat  worth  70  cents  per  bushel ;  how 
many  bushels  of  each  may  he  take  ? 

69.  A  man  owes  $6480  to  his  creditors ;  his  debts  are  in  arith- 
metical progression,  the  least  being  $40,  and  the  greatest  $500; 
required  the  number  of  creditors  and  the  common  difference  between 
the  debts.  a       ^24  creditors. 

■^^"'^-  }  $20  diilerence. 

70.  Two  ships  sail  from  the  same  port ;  one  goes  due  north  128 
miles,  and  the  other  due  east  72  miles ;  how  far  are  the  ships  from 
each  other?  Ans.    146.86 -|- 7iiiies. 

71.  If  10  pounds  of  cheese  be  equal  in  value  to  7  pounds  of 
butter,  and  11  pounds  of  butter  to  2  bushels  of  corn,  and  14  bushels 
of  corn  to  8  bushels  of  rye,  and  4  bushels  of  rye  to  1  cord  of  wood ; 
how  many  pounds  of  cheese  are  equal  in  value  to  10  cords  of  wood  ? 

Ans.    550. 

72.  A  and  B  traded  until  they  gained  6  per  cent,  on  their  stock  ; 
then  I  of  A's  gain  was  $18 ;  if  A's  stock  was  to  B's  as  f  to  ^,  how 
much  did  each  gain,  and  what  was  the  original  stock  of  each  ? 

.       3  ^'s  g'^i^i)  '^'l^  ;      stock,  $750. 
"^"^•^B's      "     $37.50;    "        $825. 

73.  If  20  men,  in  21  days,  by  working  10  hours  a  day,  can  dig  a 
trench  30  ft.  long,  15  ft.  wide,  and  12  ft.  deep,  when  the  ground  is 
called  3  degrees  of  hardness,  how  many  men,  in  25  days,  by  v\ork- 
ing  8  hours  a  day,  can  dig  another  trench  45  ft.  long,  16  ft.  wide, 
and  18  ft.  deep,  when  the  ground  is  estimated  at  5  degrees  of 
hardness  ?  Ans.    84. 

74.  Wishing  to  know  the  height  of  a  certain  steeple,  I  measured 
the  shadow  of  the  same  on  a  horizontal  plane,  27-|-  feet ;  I  then 
erected  a  10  feet  pole  on  the  same  plane,  and  it  cast  a  shadow  of  2|- 
feet ;  what  was  the  height  of  the  steeple  ?  Ans.    1031  ft, 

75.  A  can  do  a  piece  of  work  in  3  days,  B  can  do  3  times  as 
much  in  8  days,  and  C  5  times  as  much  in  12  days ;  in  what  time 
can  they  all  do  the  first  piece  of  work  ?  Ans.   ^  da. 


830  PROMISCUOUS   EXAMPLES, 

76.  A  person  sold  tMo  farms  for  $1890  eacli ;  for  one  he  roccived 
25  per  cent,  more  than  its  true  value,  and  for  the  other  2o  per  cent, 
less  than  its  true  value  ;  did  he  gain  or  lose  by  the  sale,  and  how 
much  ?  Alls.   Lost  $252. 

77.  Three  men  paid  $100  for  a  pasture;  A  put  in  9  horses,  1}  12 
cows  for  twice  the  time,  and  C  some  sheep  for  2-^  times  as  long 
as  B's  cows  ;  C  paid  one  half  the  cost ;  how  many  sheep  had  he, 
and  how  much  did  A  and  B  each  pay,  i)rovided  G  cows  eat  as  much 
as  4  horses,  and  10  sheep  as  much  as  o  cows  ?     ^  C  had  25  sheep. 

Ans.  ^  A  paid  $18. 
(B     "     $32. 

78.  A  man  purchased  goods  for  $10500,  to  be  paid  in  three  equal 
installments,  without  interest ;  the  first  in  3  months,  the  second  in 
4  months,  the  third  in  S  months ;  how  much  ready  money  will  pay 
the  debt,  money  being  worth  7  per  cent.  ?       Ans.    $10203.94 -(-• 

79.  A  farmer  sold  50  fowls,  consisting  of  geese  and  turkeys;  for 
the  geese  he  received  $.75  apiece,  and  for 'the  turkeys  $1.25  apiece, 
and  for  the  Avhole  he  received  $52.50 ;  how  many  were  there  of 
each  ?  Ajis.    20  geese,  30  turkeys. 

80.  There  is  an  island  73  miles  in  circumference,  and  3  footmen 
start  together  and  travel  around  it  in  the  same  direction  ;  A  goes  5 
miles  an  hour,  B  8,  and  C  10;  in  Avhat  time  will  they  all  come 
together  again  if  they  travel  12  hours  a  day  ?         Ans.    Gda.  1  h. 

81.  A,  B  and  C  are  to  share  $100000  in  the  proportion  of -^,  l, 
and  \,  respectively;  but  C  cl3-ing,  it  is  required  to  divide  the  whole 
sum  proportionally  between  the  other  two ;  how  much  is  each  one's 
share?  .       5  A's,  $57142.854. 

^"^•^B's,  $42857. 14f 

82.  A,  B,  and  C  have  135  sheep;  A's  plus  B's  are  to  B's  plus 
C's  as  5  to  7,  and  C's  minus  B's  to  C's  plus  B's  as  1  to  7  ;  how  many 
has  each  ?  Ans.   A,  30  ;  B,  45  ;  C,  CO. 

83.  A  man  sold  one  hog,  weighing  250  pounds,  at  4  cents  per 
pound  ;  a  second,  weighing  300  pounds,  at  4^  cents  ;  and  a  third, 
weighing  309  ])ounds,  at  5  cents ;  what  was  the  average  price  per 
jjound  for  the  whole  P  Ans.    4-|j-|  cents. 

84.  In  a  certain  factory  are  employed  men,  women  and  boys  ; 
the  boys  receive  3  cents  an  hour,  the  women  4,  and  the  men  6;  the 
boys  work  8  liours  a  day,  the  women  9,  and  the  men  12  ;  the  boys 
receive  $5  as  often  as  the  Avomen  $10,  and  for  every  $10  ))aid  to  tlie 
women,  $24  are  paid  to  the  men ;  how  many  men,  women,  and 
boys  are  there,  the  whole  number  being  59  ? 

Jns.    24  men,  20  women,  15  boys. 

85.  A  fountain  has  4  receiving  pipes.  A,  B,  C,  and  ]) ;  A,  B,  and 
C  will  fill  it  in  G  hours,  B,  C,  and  ])  in  8  hours,  C,  D,  and  A  in  10 
hours,  and  D,  A,  and  B  in  12  hours;  it  has  also  4  discharging 
pipes,  W,  X,  Y,  and  Z  ;  AV,  X,  and  Y  will  empty  it  in  G  hours,  X,  Y, 
and  Z  in  5  hours,  Y,  Z,  and  "W  in  4  liours,  and  Z,  "W.  and  X  in  3 
hours  ;  suppose  the  pipes  all  open,  and  the  fountain  full,  in  what 
time  would  it  be  emptied  ?  -^I;;*'.   Gj'L  h. 


PROMISCUOUS    EXAMPLES.  331 

86.  How  many  building  lots,  each  75  feet  by  125  feet,  can  be 
laid  out  on  1  A.  1  11.  6  P.  18^-  sq.  yd.  ?  Jns.   6. 

87.  A  man  bought  a  house,  and  agreed  to  pay  for  it  $1  on  the 
first  day  of  January,  $2  on  the  hrst  day  of  February,  $4  on  the 
first  day  ]\lareh,  and  so  on,  in  geometrical  progression,  through 
the  year  ;  what  was  the  cost  of  the  house,  and  what  the  average 
time  of  payment  ?  .       ^$1095. 

■  (  Average  time,  Nov.  1. 

88.  A  man  sold  a  rectangular  piece  of  ground,  measuring  44 
chains  32  links  long  by  3G  chains  wide ;  how  many  acres  did  it 
contain  ?^  Ans.    159  A.  2  11.  8.32  P. 

80.  What  number  is  that  which  being  increased  by  its  half,  its 
tliird,  and  18  more,  M-ill  be  doubled  ?  Ans.    108. 

90.  A  merchant  has  200  lb.  of  tea,  worth  $.62^  per  pound,  which 
he  will  sell  at  $.56  per  pound,  provided  the  purchaser  will  pay  in 
coflee  at  22  cents,  which  is  Morth  25  cents  per  pound ;  does  the 
merchant  gain  or  lose  by  the  sale  of  the  tea,  and  how  much  per 
cent.  ?  Ans.    gained  l-fj  per  cent. 

01.  A  man  owes  a  debt  to  be  paid  in  4  equal  installments  at  4, 
9,  12,  and  20  months,  respectively;  discount  being  allowed  at  5  per 
cent.,  he  finds  that  $750  ready  money  will  pay  the  debt;  how  much 
did  he  owe  .P  Ans.    $784.74-}-. 

92.  A  and  I>  traded  upon  equal  capitals  ;  A  gained  a  sum  equal 
to  "I  of  his  capital,  and  B  a  sum  equal  to  -IL  of  his;  B's  gain  was 
$500  less  than  A's  ;  what  was  the  capital  of  each  ?     Ans.  $4000. 

93.  I  purchase  goods  in  bills  as  follows:  June  4,  1859,  $240.75  ; 
Aug.  9,  1859,  $137.25;  Aug.  29,  1859,  $05.04;  Sept.  4,  1S59, 
$230.36;  Nov.  12,  1859,  $36.  If  the  merchant  agree  to  allow 
credit  of  6  mo.  on  each  bill,  Avhen  may  I  settle  by  paying  the  whole 
amount?  "  Ails.    Feb.  1,  1860. 

94.  A  young  man  inherited  a  fortune,  i  of  which  he  spent  in  3 
months,  and  4  of  the  remainder  in  10  months,  when  he  had  only 
$2524  left ;  how  much  had  he  at  first  ?  Ans.  $5889.33  -[-. 

95.  A  man  bought  a  piece  of  land  for  $3000,  agreeing  to  pay  7 
per  cent,  interest,  and  to  pay  princij^al  and  interest  in  5  equal  au- 
r^ual  installments  ;  how  much  was  the  annual  payment  ? 

^4)?.'?.   $731.67  -[-. 
D6.  I  have  three  notes  payable  as  follows :  one  for  |200,  due 
Jan.  1.  1859,  another  for  $350,  due  Sept.  1,  and  another  for  $500, 
due  April   1,    1860  ;  what  is  the  average  of  maturity  ? 

A71S.  Oct.  24,  1850. 
97.  A  man  held  three  notes,  the  first  for  $600,  due  July  7,  1859; 
the  second  for  $530,  due  Oct.  4,  1859 ;  and  the  third  for  $400,  due 
Feb.  20,  1860  ;  he  made  an  equitable  exchange  of  these  Milh  a 
s])eculator  for  two  other  notes,  one  of  Mhich  was  for  $730,  due  Nov. 
15,  1859;  what  was  the  face  of  the  other,  and  when  due? 

.       5  Face,  $800. 
^"*-  )  Due  Aug.  29,  1859. 


332  MENSURATION. 

ilEXSURATIOX    OF   LINES  AND   SUPERFICIES. 

449'.  In  taking  the  measure  of  any  line,  surface,  or  solid,  we  are 
always  o-overncd  by  some  denomination,  a  unit  of  wliich  is  called  the 
Unit  of  Measure.  Thus,  if  any  lineal  measure  be  estimated  in  feet, 
the  unit  of  measure  is  1  foot ;  if  in  inches,  the  unit  is  1  inch.  If  any 
superficial  measure  be  estimated  in  feet,  the  unit  of  measure  is  1 
square  foot ;  if  in  yards,  the  unit  is   1  square  yard. 

44§.  If  any  solid  or  cubic  measure-  be  estimated  in  feet,  the 
unit  of  measure  is  1  cubic  foot ;  if  in  yards,  the  unit  is  1  cubic  yard. 

419.   The  area  of  a  figure  is   its    superficial  contents,  or  the 
surface  included  within  any  given  lines, 
without  regard  to  thickness. 

430.  An  Oblique  Angle  is  an  angle 
greater  or  less  than  a  right  angle ;  thus, 
ABC  and  C  B  D  are  oblique  angles, 

CASE    I. 

451.    To  find  the  area  of  a  square  or  a  rectangle. 

4.'52.  A  Square  is  a  figure  having  four  equal  sides  and  four 
right  angles. 

453.  A  Kectaugle  is  a  figure  having  four  right  angles-,  and 
its  opposite  sides  equal. 

Rule.  Multiply  the  length  ly  the  breadth,  and  the  product  icill 
be  the  square  contents. 

EXAMPLES    FOR    PKACTICE. 

1.  IIov.-  many  square  inches  in  a  board  3  feet  long  and  20  inches 
wide?  "  ^ns._  720. 

2.  A  man  bought  a  farm  198  rods  long  and  loO  rods  wide,  and 
af^reed  to  give  $32  an  acre  ;  how  much  did  the  farm  cost  him  ? 

An^.   85940. 

3.  A  certain  rectangular  piece  of  land  measures  1000  links  by 
100  ;  how  many  acres  does  it  contain  ?  Ans.  1  A. 

CASE    II. 

^5S.    To  find  the  area  of  a  rhombus  or  a  rhomboid. 

455.  A  SllOinbus  is  a  figure  having  four  equal  sides  and  four 
oblique  angles. 

450.  A  Ehoniboid  is  a  figure  having  its  opposite  sides  equal 
and  parallel,  and  its  angles  oblique. 

Note.  The  square,  rectanp;lc,  rhombus,  and  rhomlioid,  havinp;  their  op- 
posite sides  parallel,  are  called  by  the  general  name,  parcdlehxjram. 

It  is  proved  in  geometry  that  any  parallelogram  is  equal  to  a  rec- 
tangle of  the  same  length  and  width ;  hence  the 


MENSURATION.  333 

Rule.  Multiply  the  lengtJi  by  tJie  shortest  or  perpendicular  dis- 
tance between  tioo  opposite  sides. 

EXAMPLES    FOR    rUACTICE. 

1,  A  meadow  in  the  form  of  a  rhomboid  is  20  chains  long,  and 
the  shortest  distance  between  its  longest  sides  is  12  chains  ;  how 
many  days  of  10  hours  each  will  it  take  a  man  to  mow  the  grass  on 
this  meadow,  at  the  rate  of  1  square  rod  a  minute  ?     Ans.  6  da.  4  h. 

2.  Tlic  side  of  a  plat  in  the  form  of  a  rhombus  is  15  feet,  and  a 
perpendicular  drawn  from  one  oblique  angle  to  the  side  opposite, 
will  meet  tliis  side  9  feet  from  the  adjacent  angle ;  what  is  the  area 
of  the  plat  ?  Ans.    180  sq.  ft. 

CASE   III. 

457.     To  find  the  area  of  a  trapezoid. 

45§.  A  Trapezoid  is  a  figure  having 
four  sides,  of  wliich  two  are  parallel. 

The  mean  length  of  a  trapezoid  is  one 
half  the  sum  of  the.  parallel  sides ;  hence  the 


PtULE.  Multiply  one  lialftlie  sum  of  the  parallel  sides  by  the  per- 
ITcndicular  distance  between  them. 

EXAMPLES    FOR    PRACTICE. 

1.  "What  are  the  square  contents  of  a  board  12  feet  long,  16 
inches  wide  at  one  end,  and  9  at  the  other  ?  Ans.    12-^  sq.  ft. 

2.  "^Vhat  is  the  area  of  a  board  8  feet  long,  16  inches  wide  at 
each  end,  and  8  in  the  middle  ?  Ans.    8  sq.  ft. 

0.  One  side  of  a  field  is  40  chains  long,  the  side  parallel  to  it  is 
22  chains,  and  the  perpendicular  distance  between  these  two  sides 
is  2o  chains  ;  how  many  acres  in  the  field  ?      Ans.  77  A.  5  sq.  ch. 

CASE    IV. 

459,    To  find  the  area  of  a  triangle. 

450.  The  Base  of  a  triangle  is  the  side  on  which  it  is  supposed 
to  stand. 

461.  The  Altitude  of  a  triangle  is  the  perpendicular  distance 
from  the  angle  opposite  the  base  to  the  base,  or  to  the  base  produced 
or  extended. 

462.  A  Triangle  is  one  half  of  a  parallelogram  of  the  same 
base  and  altitude ;  hence  the 

Rule.  Mvltiply  one  half  the  base  by  the  altitude,  or  one  half  the 
altitude  by  the  base.  Or/ Midtiply  the  base  by  the  altitude,  and 
divide  the  product  by  2. 

EXAMPLES    FOR    PRACTICE. 

1.  How  many  square  yards  in  a  triangle  whose  base  is  14S  feet, 
and  perpendicular  45  feet?  Ans.  370yds. 


334  MENSURATION. 

2.  The  gable  ends  of  a  barn  are  each  28  feet  wide,  and  tlie  per- 
pendicular height  of  the  ridge  above  the  eaves  is  7  feet ;  how  many 
feet  of  boards  will  be  required  to  board  up  both  gables  P 

Alls..  IDG  feet. 

CASE    V. 

46.3.     To  find  the  circumference  or  the  diameter  of  a  circle. 

464.  A  Circle  is  a  figure  bounded  by  one 
miiform  curved  line. 

46.5.  The  Circumference  of  a  circle  is  the 
curved  line  bounding  it. 

466.  The  Diameter  of  a  circle  is  a  straight 
line  passing  through  the  center,  and  termina- 
ting in  the  circumference. 

It  is  proved  in  geometry  that  in  every  circle  the  ratio  between  the 
diameter  and  the  circumference  is  3.1416  -)-.     Hence  the 

Rule.  I.  To  find  the  circumference.  —  Mult!pJ>j  the  diamefer 
by  3.1416. 

II.    To  find  the  diameter.  —  Multiply  the  circumference  by  .3183. 

EXAMPLES    FOR    PRACTICE. 

1.  What  length  of  tire  will  it  take  to  band  a  carriage  wheel  5 
feet  in  diameter  ?  Ans.    15  ft.  8.4  -|-  in. 

2.  AVhat  is  the  circumference  of  a  circular  lake  721  rods  in 
diameter  ?  Ans.    7  mi.  25  rds.  1.54  -\-  ft. 

3.  Vrhat  is  the  diameter  of  a  circle  33  yards  in  circumference  ? 

Ans.    10.5  -\-  yards. 

CASE   VI. 

467.  To  find  flie  area  of  a  circle. 

From  the  principles  of  geometry  is  derived  the  following 

Rule.  I.  When  both  diameter  and  circumference  are  given  ;  — 
Mnltiplij  the  diameter  by  the  circumference,  and  divide  the  inoduct 
&y4.  _ 

II.  AVTien  the  diameter  is  given;  —  Multiply  the  square  of  the 
diameter  by  .7854. 

III.  When  the  circumference  is  given  ;  —  Multiply  the  square  of 
the  circunij'erence  by  .07958. 

EXAMPLES    FOR    PRACTICE. 

1.  The  diameter  of  a  circle  is  113,  and  the  circumference  355  ; 
Avhat  is  the  area  ?  Ans.    10028.75. 

2.  What  is  the  diameter  of  a  circular  island  containing  1  square 
mile  of  land  ?  Ans.  1  mi.  41  rd.  1.4 -F  ft. 

3.  A  man  has  a  circular  garden  requiring  84  rods  of  fencing  to 
inclose  it;  huw  nuich  laud  in  the  garden  ?      Aiis.  3  A.  81.5-}- 1*. 


I 


MENSURATION.  335 


JiIENSURATIOX   OF    SOLIDS. 

468.     A   Solid   or  Body  is   a   magnitude  which  has  lengtli, 
breadth,  and  thickness. 

CASE    I. 

4G9.  To  find  the  cubic  contents  of  a  prism, 
cube,  or  cylinder. 

470.  A  Prism,  is  a  solid  whn.se  bases  or  ends 
are  any  similar,  equal,  and  parallel  plane  figures, 
and  whose  sides  are  parallelograms. 


4:71.   A  Cylinder  is  a  body  whose  bases     ^g^g 
or  ends   are    equal   and  parallel  circles,   and  ^ 

whose  side  is  a  uniform  curved  surface. 


4'y2.  The  Altitude  of  a  prism,  cube,  or  cylinder,  is  the  perpen- 
dicular distance  between  the  two  bases  ;  it  is  the  leiujth  of  the  body. 

To  estimate  the  solid  contents  of  any  one  of  the  bodies  defined 
under  this  case 

Rule.     Multiphj  the  area  of  the  base  hy  the  altitude. 

EXAMPLES    FOR    rUACTICE. 

1.  The  side  of  a  cubic  block  measures  8  inches  ;  hoAV  many  cubic 
inches  does  it  contain  ?  Ans.    512. 

2.  The  end  of  a  prism  20  feet  long  is  a  right-angled  triangle,  the 
two  shorter  sides  of  wliich  measure  9  and  12  inches  ;  what  are  the 
cubic  contents  of  the  prism  ?  Ans.    1\  cu.  ft. 

3.  A  stick  of  timber  is  2-3  ft.  3  in.  long,  1  ft.  8  in.  wide,  and  18  in. 
thick ;  how  much  will  it  come  to  at  8  cents  per  cubic  foot  ? 

Ans.    $5.05. 

4.  A  cistern  is  5\  feet  in  diameter,  and  8  feet  deep  ;  how  many 
standard  wine  gallons  will  it  contain  ?  ^4??.?.    1421.7984  gal. 

Notes.  1.  The  mean  or  average  diameter  of  a  barrel  or  cask  may  be 
found,  by  adding  to  the  head  diameter  2,  or,  if  the  staves  be  but  little  curv- 
ing, Yo  °^  t'^^  difference  between  the  head  and  bung  diameters.  The  cask 
will  then  be  reduced  to  a  cylinder,  and  its  contents  found  by  the  above  rule. 

2.  The  process  of  estimating  the  capacity  of  barrels  or  casks  is  called 
gaKging. 

0.  The  head  diameter  of  a  cask  is  22  inches,  the  bung  diameter 
28  inches,  and  the  length  31  inches ;  how  many  wine  gallons  will  it 
contain?  Ans.    71.2504. 

6.  The  head  diameter  of  a  cask  is  30  inches,  the  bung  diameter 
35  inches,  and  the  length  40  inches  ;  what  is  its  capacity  ? 


335 


MENSURATION. 


CASE   II. 

473.  To  find  the  cubic  contents  of  a  pyramid 
or  a  cone. 

474.  A  Pyramid  is  a  solid  whose  base,  is  any 
plane  figure,  and  whose  sides  are  triangles   terminat-  / 
ing  in  a  point  at  the  top. 


^75.  x\  Cone  is  a  solid  whose  base  is  a  circle, 
and  whose  side  is  a  curved  surface  terminating  in  a 
point  at  the  tojj. 

Rule.  Multiply  the  area  of  the  base  hy  \  of  the 
altitude. 

EXAMPLES    FOR    PRACTICE. 

1.  What  are  the  solid  contents  of  a  pyramid  15  feet  square  at 
the  base  and  40  feet  high  ?  Ans.  3000  cu.  ft. 

2.  A  pyramid  has  a  triangular  base,  each  side  of  Avhich  is  30 
inches,  and  the  altitude  of  the  pyramid  is  4  feet ;  what  are  the  cubic 
contents  ?  Ans.   3.6  -f-  cu.  ft. 

3.  The  base  of  a  cone  is  7  feet  in  diameter,  and  the  altitude  IG 
feet  9  inches  ;  what  are  the  solid  contents  ?   Ans.    214.87  -|-_cu.  ft. 

4.  A  heap  of  grain,  in  the  form  of  a  cone,  is  4  feet  high,  and 
measures  15  feet  round  the  base  ;  how  many  bushels  does  it  con- 
tain? Ans.    19  bu.  5.9-j- qt. 

CASE  III. 

476.  To  find  the  surfoce  or  the  solid  contents  of  a  sphere. 

477.  A  Sphere  or  Glohe  is  a  solid  bound- 
ed by  a  single  curved  surface,  which  m  every 
part  is  equally  distant  from  a  point  within  called 
Its  center.    Hence 

Rule.  I.  To  find  the  surface ;  —Multiply  the 
square  of  the  diameter  hy  3.1416. 

II.    To  find  the  solid  contents  ;  —  Midtiply  the 

cube  of  the  diameter  by  .5236. 

EXAMPLES    FOR   PRACTICE. 

1 .  How  many  square  inches  on  the  surface  of  a  globe  1 5  inches 
in  diameter?  yl».9.  706.S6. 

2.  The  diameter  of  a  sphere  is  18  inches ;  what  is  its  solidity? 

3.  "What  is  the  solidity  of  a  ball  that  can  just  be  put  into  a  cylin- 
drical cup  5  inches  in  diameter  and  5  inches  deep  ? 

Ans.  65.45  cu.  in. 


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A    DESCRIPTIVE     CAT.i.  -  B.     containing    Full     Bcscriptions, 

Terms,  Ciitio;il  Notices,  C'oiiiiiie..  i.  m.s  ■  i  Tenclicrs.  etc.,  of  tlioir  Scrios 
of  Koiwlers,  Histories,  Natural  =.ioii.c  n  1  all  tluir  Publications,  sent, 
prp-i.ai<l,  on  application.  Hln?U  vopi.  .  ■  i  tl.rt  above  books  sent,  prc-jinid, 
by  mail  to  any  ruldrcss.  on  rcc  iit  o."  i'.'  retail  price,  and  of  those  use.l  In 
clas-tc-s,  TO  TKAOHERrf,  at  fiul.'  rfl'ii  r«rk\;.  for  examination.  Specialbj 
favorable  termn  for  first  introdi'  •Hoi 

IVISON,  PHINN.GT,  .     AKEMAN  &  Co., 

48  fi.    .c  Walker  Street,  N.  Y. 


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